Transcript for:
Understanding 1D Motion in Physics

[Music] [Music] [Music] [Music] [Music] [Music] [Music] [Music] [Music] [Music] [Music] how are you all so uh this was a bike sequence regarding right explaining the concepts of motion in uh one day yes uh we are talking about the kinematics right we have seen the pw headquarters okay so guys this thing this kind of things will always come upon right so show your support towards this lecture especially for the teacher and sharing maximum if you really really really liked it okay guys sorry for this stuff so this is your own physics educator indrajit singh welcoming you all into the most sensible platform for an experience those who are preparing for jai need kind of examination and a very warm welcome for all right into the victory badge this is a one shot lecture on the motion into the one day okay lots of concepts we are going to see within it lots of concept right we are going to see within it what kind of the concepts right what kind of the concepts lots of things are going to come right see before moving on to this part i want to tell you all about uh some things we are going to deal with just hold on the fundamentals part the uh see basically this all chapter is entirely about the three kind of motion first one is uniform second that is a uniformly accelerated motion and lastly it is a non-uniform motion but there are some what you can say the least cases of the non-uniform motion and apart from that right one uh concept that is common in 1d and 2d that is relative motion okay so this thing we are going to say in today's lecture okay uh fundamentals uniform motion uniformly accelerated motion graph based on this concept okay and the concepts of relative motion into the one day graft based on this concept specially we are going to have right 30 to 45 minutes for the uh to understand the concept of graphs and their respective graph conversion the je net exam i'm sorry for this the j and i'm sorry extremely sorry okay the je and nate examination okay holds the prominent position okay even this topic if i'm talking about the graph conversion right then the holds high weightage and examination and having the highest possibility to come up in to the paper right so this kind of uh things we are going to discuss it today so shall we begin yes or no shall we begin okay so shall we begin okay so let's start with uh see the chapter starts with a very simple concept right the concept of motion what does it states okay and before moving on to the motion i want to tell you all one thing okay from now onwards still gravitation like 1d 2d nlm work energy and power and slowly slowly we'll move on to the rotation and gravitation till gravitation we are going to deal with the two branches of the physics we are going to deal with the two branches of the physics the one that is kinematics the second one that is the dynamics so we are going to deal with the two branches right this is the first one that is the kinematics second that is a dynamics kinematics if i am talking about see this entire chapter is kinematics it is the study of the motion right without cause with cause without cause no investment no involvement of force no involvement of energy right okay nothing no involvement of friction no resistive no dissipative forces forces i'm talking about right so this entire chapter is completely based on the kinematics kinematics in another sense everything will be without cause i am the observer you are the observer you are analyzing the motion and everything it is going without cause we are just analyzing in the question if data is given object is moving from this point to this point yes we are accepting right we are not right we are not getting worried about it we are not understanding what is a logic behind it yes we will uh correct the logic behind solving these sums right but we are actually not dealing with the force okay so there is a difference this difference you should all be aware of at this point of time you should be aware of okay so basically kinematics without cause dynamics with cause with cause it means we are going to deal with the force right we are going to deal with the energies right okay the work energy theorem and lots of things in another sense we are going to deal with the we are going to study the cause of the motion in its respective applications right the cause of the motion and its respective applications so this is uh the very basic thing that i want to discuss you with you all now let's move on to so in this entire chapter okay at any point of time if any student is having the doubt in their mind why this sort of thing is going on why acceleration is acting like this way y value of acceleration is getting changed so the answer for all of those doubts is only one and that is we are dealing with without cause we are only analyzing the motion so in that case whatever the parameters whatever the situation it is given we have to accept it and we have to try to find out the way to reach out the solutions into the numericals okay so what is a motion right what actual is a motion is this is a day-to-day uh right example car on a road is it a motion sir it happens it completely depends whether you are inside the car or whether you are standing on the ground analyzing the moving car or sometimes it also depends on what thing it depends students might be saying yes the car is on the road i am saying the car is moving on the road right see this is having the two meanings car is on the road so is it stationary on the road or is it moving for a person who is standing on the ground there is a possibility that car may move with some specific velocity but right for the same person who is inside the car right okay he appears to be it appears the car appears to be at rest with respect to that post so for someone car is in motion for someone car is at rest so practically we can't say anything why we can't say anything because everything depends on what observer is observing in the phenomena again and again i'm saying mark my words everything depends on what observer is observing in the phenomena but if we want to in generalized form if we want to from the kinematics point of view if you want to define the motion then we can say that right change in position of an object with respect to time and actually your word observer should surrounding should come but i am saying observer right i am saying it an observer why because this hold the most prominent position everything is with respect to observer and with respect to time everything is with respect to each and every motion is analyzed with respect to observer okay observer changes things may get changed observer position changes great so here i gave you all an example the car is on the road with respect to the person who is inside right it's stationary okay for him the surround the surrounding environment is moving but the person who is on the ground is analyzing the motion for him the car is moving so we are with the two observer no the first observer is on the ground the second observer is inside it the car so things always get differs if the position of the observer gets changed if the in another sense if i want to note it down you want to deliver it technically if the frame is getting changed things are getting changed the frame is getting changed the things are getting changed so it everything depends on frame of reference right everything depends on the frame of reference what actual the frame of reference is right what actual the frame of reference is lots of technical definitions are given into the thousands of the books but i want to deliver it into the right i want to deliver it specifically okay and i always used to say in all of my sessions that whatever the things that are easily understood by you all are always represented rather than okay rather than to move on to the quite complex approach these things whatever the definition i'm going to write within few minutes it would be quite easily understood by you all apart from the essay understanding the concept it is important okay rather than to move on to the grammatical formations right if it is in your mind again you can't even write it down and you can even uh you will not face any kind of the problem while analyzing the multiple resources right so frame of references frame of references so i'm going to simply define this term the frame of references i'm going to define it as the point from which observer takes its observation this point can be any position in 1d right can be in 2d can be any 3d coordinates right this point can be any point at any position into the entire three-dimensional space again we are coming back to our same example a car on the road right a one point is the person who is standing on the ground and analyzing the motion the second point is the person who is actually inside the car things may get different right so the frame of reference the point from which observer takes its observation simply as i said i'm not having not given you that much technical approach but simple it is needed frame is getting changed things are getting changed motion is not getting changed motion will get changed technically because things may get different and according to that you are going to apply those specific concepts into the numericals right so frame of reference and one thing i want to write it down here right and that is nothing in a universe is in absolute motion while nothing is at absolute rest is earth appear the the earth appears stationary to you all why because you all are inside the frame of reference with respect to moon earth is not stationary with respect to sun earth is not stationary with respect to mass earth is not stationary right even the situations are not same while analyzing the earth from mars and while another analyzing the earth from saturn or bluetooth right okay so everything is a frame of reference nothing in the universe is at absolute rest we can't define that from all frame of references see absolute rest it means that in the entire universe right from all frame of references from all positions right it appears to be stationary is it possible no is it not it's not possible right so for that reason nothing in the universe is at absolute rest similarly absolute motion it means that in the universe that exists all coordinates all points right from which the observer is observing and they are getting the same results same result it means the person who is standing on the road analyzing the car is also being the same output as compared to the person who is like inside it and drive you can see the inside it and it is driving the car is it so possible no so nothing in the universe is at absolute rest right nothing in the universe is an absolute motion okay now basically we are having the two types of frame of references the first one that is an inertial frame of reference right the second one that is in a non-inertial frame of reference now before moving on to before moving on to this concept inertial and non-inertial frame i want to clarify write a thing that from now onwards whatever you are going to analyze you are going to be the observer why you are all going to be the observer because things are going to appear in front of you all you are going to analyze these things it means you are the observer so whatever the frame right you are holding okay that will be the truth you have to analyze all of those concepts through that frame okay you have to analyze all of those concepts to that frame inertial and non-inertial frame of reference right so from now onwards observer is in the specific frame right when right when the frame velocity is zero or when the frame velocity is zero or the observer is moving with the constant velocity when the frame velocity is zero or the observer is moving with the constant velocity right then we can say that what is an example of it okay this is a cart it is moving with some right velocity let's say it's constant but here this is an observer this is the position of the observer observer is stationary velocity of the observer is zero then i must say he is in the inertial frame second cart is moving with some value ct v1 observer is also moving with some value cd v2 v1 is also constant v2 is also constant so in that case this frame is also inertial frame right values may get differ right outputs may get different values may get different outputs may get differ right but the one is an inertial frame right the frame velocity or the velocity of an observer is zero is stationary or the velocity of the observer is constant let's move on to the next part onto it right the second this is an inertial frame of reference now here let's move on to the non-inertial frame of reference when the frame velocity is continuously increasing or decreasing right frame velocity is continuously increasing or decreasing with respect to time continuously it is getting increasing or decreasing with respect to time let's analyze this stuff right of the non-inertial frame right let's analyze the stuff of the non-inertial frame okay the cart is moving with the constant velocity but the observer who is on the ground right his velocity is continuously increasing with respect to time right observer velocity increases with respect to time inertial frame of reference non-inertial frame of reference right so this was all of those basics of the right the motion okay now sign convention see the this is a very problematic case because in some of the schools right they are taking the downward direction as a positive right in upward direction as a negative and while in some schools right they are taking upward as a positive and downward is a negative i'm going to tell you all the general sign conventions that we are going to use till end of this course this end of this victory batch end of this victory batch till end we are going to use that stuff sign conventions right this is x-axis this is y-axis this is z-axis the same thing we are going to use till end right so remember vectors you all are aware of vectors you all are aware of vectors means a very simple thing suppose a vector of a physical quantity is directed upwards we are going to take it as a positive j so it means upward will be positive if vector of any physical quantity is directed downwards we are going to take it into the negative j it means whatever the vector right that are directing into the downward directions are considered as negative the vector that is directed towards the positive x direction right i will consider it as a positive i it means whatever the vectors are directing towards the right will be positive right and the vector that is right moving into the right that is directed towards the negative x direction right i will consider it as the uh the negative i right so whatever the vectors they are directed into the negative direction towards this side they will consider as a negative similarly that exists for the z also this side it is positive and the opposite side this will negative but here right now we are only dealing with the motion in one day uh in the kind of the rectangular the rectilinear motion we are dealing with right okay so we are only moving on to we will only consider x and y upward positive downward negative okay so in some of the schools it is this skydive some of the right institutions it is they can always use to they always used to take downward as positive but just keep this thing women everything is with respect to the reference point right okay without a reference point we can't analyze the motion okay as uh like things always get changed with respect to the observer similarly if the reference point is getting changed things always the parameters of the motion always gets changed the reference point is getting changed the parameters of the motion always gets changed right so everything is with respect to the reference point and the reference point of the motion is t is equal to 0. generally in maximum number of the cases when observer starts analyzing the motion we can say it's t is equal to zero yes i started he might have started from mumbai but i met him at suru right so t is equal to zero i have met him this is a it is possible that object might be running since mumbai and even he had covered or the object is covered distance also but i started analyzing from surat na so for that observer it is t is equal to 0 there is a possibility that you may start from t is equal to 0 but observers may get different so what observer is thinking that the moment at which observer tries to write analyze starts to analyze the motion things are fixed it is a reference point and that reference point is always considered as t is equal to zero right so for the sign conventions right it is cleared for you all from now onwards we are going to take upward as the positive this are the generalized sign conventions that we are going to use till end of the physics apart from this okay two more sign conventions will come one will come into the rotational motion clockwise and anti-clockwise the second will come up into the magnetism into the twelfth part right the inward and outward but that invert and outward is also part of the z axis right this is the third dimension these add x's okay so only apart from this sign conventions only two new things are going to come one will come in rotation and second will come into the slivers of 12. right so these are the sign conventions let's move on to the next part the fundamentals okay position now the motion starts right as i used to say lots of technical definitions of the positions are mentioned okay but i want to move on to the very simple thing i want to move on to the very very simple thing right what is the position okay at a specific the one who understand this statement will get the definition of the position at a specific time instance where the object is at the specific time instance where the object is at t is equal to zero i started my motion with respect to some observer from home at a is equal to zero i started the motion with respect to some observer on home from home slowly slowly slowly it started right so from t is equal to zero it started what is the i want to reach up to the office it will take approximately one minute let's say one minute what is the position of the object at t is equal to five seconds what is the like i can't say what is the position right if i starts from t is equal to zero right from home then i might say where i reached at t is equal to phi first of all t is equal to zero will be my reference point this may get changed but generally maximum number of the question t is equal to 0 is taken as the reference point there exists one question in ncrt right in which the reference point has changed okay time is considered as the negative okay but that thing is conceptually correct also don't get worried okay but maximum number of the cases says this so from k is equal to zero right this is a reference point and things may get stuck right now i'm in the continuous call with my colleague after five seconds where are you oh yeah i have crossed the 7th street after 25c where are you i've crossed this blabla signal after 40 seconds where are you after 45 where are you after 55 where are you after 60 in front of you so at a specific time instance what is the position what at what specific time instance where object is will define the position of the object with respect to the reference point at t is equal to 5 this is the position of the car at t is equal to 20 this will be the position of the car at t is equal to 30 this will be the position of the car at 3 is equal to 40 this will be the position at t is equal to 50 this will be the position so at the specific time instance right where actually the object is located with respect to an observer at specific time instance right actually where object is located with respect to the specific time instance will define the position of the object right this in one day it is a position in 2d it is a coordinate right okay when object is moving into the plane okay this will come up into the next chapter right okay now let's move on to the next after this position now we are dealing with the parameters of the motion the first parameter is a position the second is the path length now what actual the path length is all about what actual the path length is all about right path length first of all let's understand the example of the path length at t is equal to zero this is our reference point from which right observer or uh okay object start its motion and observer is analyzing suppose here is an observer okay t is equal to zero he starts analyzing the motion at t is equal to zero he starts analyzing the motion what is the path length it doesn't matters how you are going suppose he starts from a it reaches to b then he moves on to the c then he against move on to d at t is equal to zero motion started at t is equal to d motion ends so what does this path length states the total distance covered into the motion is defined as path length the total distance covered into the motion it is defined as the path length so here what will be the right value of the path length in another sense a of b plus b of c plus c of d now students are in the cure sir what is the actual difference between the path length and the distance this you told us now the point from where like the point from which motion begins to the point it ends the point from where the motion begins to the point where it ends the total distance cover maybe it is on the circular path we are going to consider as the distance only okay and that thing will get clear while solving up the sums okay that thing will be clear by solving up these sums okay so at t is equal to zero motion started at t is equal to zero the motion started that until now until end of the motion the total amount of the path distance covered now what is the distance and what is the difference between the path length and the distance right what is the okay see this is the path length now what is a distance distance can be on between the any two time instances right distance can be in between while path length is for entire motion distance is in between the two points while the path length is for the entire motion what is the distance covered between t is equal to 0 to 2 is equal to t1 the two time instances i must say the distance is a b this is not the path length path length is entire distance covered into the journey and distance is what okay it is a specific it can be in between the two time intervals right it can be for the entire journey right it completely depends so this is the actual difference right yes i'm talking about the difference between the distance and the path length a micro thing that might not have delivered right but i'm telling you in front of generally the displacement it is the shortest path right the displacement is the shortest path examples point a point b if you are moving on to the curve right a b that is an arc a b that is an r i can consider it as a distance i'm not saying it is a path length it is only the distance right but if i'm talking about the displacement the straight line is the displacement the straight line is what the displacement is right let's see some of the sums right based on the distance and the displacement to clear out the concept right a man goes 10 meter towards north then 20 meter towards east then the displacement is what displacement is the shortest distance in between the two time intervals or in between the initial and the final point of the motion displacement is the shortest distance between the two point uh right or between the starting and the end point of the motion yes or no okay so here let's move on and let's analyze this thing i'm having the isometric view this is north this is south this is east and this is west a man goes 10 10 meter towards the north then 20 meters towards the 8th so whenever you get confused again draw the convention here this is north this is south this is east and this is west now he is moving right 20 meter towards the east this is a 20 meter so this is the final point of the motion this is the initial point of the motion right okay this is a displacement the displacement is what directly if we are going to apply the pythagoras theorem and it is under root of 20 square plus 10 square this will be the displacement right that is is equal to under root of 500 so that is equal to under root of 10 root phi and this is nearly equivalent to 22.5 meter what is the distance covered 10 plus 20 30 what is the displacement 22.5 so this is very easy huh it's not fair a body moves let's come to the fair a body moves on one fourth of the circular arc in a circle of radius r the magnitude of the distance traveled and the displacement will be what body goes over one fourth of the circular arc let's say this is an initial position and this one is a final this is a circular arc right first of all if i'm talking about the distance so that will be arc length that will be arc led for the quarter part of the circle see it is defined as theta dot r where r is the radius for quarter part of the circle the theta is defined as pi by 2 into r so this will be the distance covered right this will be the distance covered what is the magnitude of the displacement shortest distance between a and b r r root r squared r square r root right so distance travel is pi r by 2 displacement is r root 2 the a is the perfect answer so quite simple sum you are not having not showing us the complex one the displacement of a point of a wheel initially in contact with the ground when the wheel rolls forward half of the revolution will be what point is at this position initially finally after half rotation half of the rotation it means write the arc length cover the total distance covered is pi r why it is theta into r for half of the rotation theta will be considered as pipe right so that's why this linear distance covered between the two points into the half i am talking about the distance not path length right how we are going to solve this sum okay again uh to solve it up i'm giving you all the 10 10 seconds right okay on the countdown of 10 9 8 7 six five four three two one that's it okay you can pause the video and answer this specific question this is the radius 2r this is the distance covered by r can we say this is the between initial and final point this is the displacement can we apply the pythagoras theorem right on applying the pythagoras theorem finally we are going to get the displacement so the displacement that is is equal to right okay under root of pi r whole square plus 2 r whole square so this is under root of pi square r square plus 4 r square so this is r under root of pi square plus 4 this is the displacement on completing the half revolution between the two points on completing the half revolution between the two points right let's move on to the next parameter of the portion see six of the parameters we are going to discuss speed velocity average speed average value ct instantaneous speed and instantaneous velocity okay rather than to have the right individual discussion on the specific motion parameter okay i'm going to show you all the three differences the first will be the difference in between the speed and the velocity the second will be the difference in between the uh this one the speed and velocity the second will be in difference between the average speed and average velocity and the third one in between the instantaneous speed and instantaneous fellows okay to have the individual discussion on that specific motion parameter is going to consume the lots of time rather than okay this differences are going to help us a lot speed the speed velocity sorry it is a distance covered by an object per unit time it is always positive or it can be zero distance covered in a specific time distance covered into the specific time while the velocity is displacement into this specific unit time this can be positive this can be zero this can be negative speed is always positive velocity can be positive velocity can be 0 velocity can be negative it completely depends on right as i told you all into the distance distance is always positive right okay distance is always positive displacement right can be positive can be negative can be zero similarly in basis of this data right okay on basis of this data speed is always positive s i unit is meter per second si unit is meter per second having same dimensional formula speed is associated with distance so as it for those reason it is the scalar quantity but velocity is associated with the shortest distance speed is associated with the distance but the velocity is associated with the shortest distance so this shortest distance can be at any any any it can direct in any way in the entire three dimensional space it can direct in any way in entire three-dimensional space right okay so for that reason it is a vector quantity okay so using this three to four terms it is easily right it is we can easily define now what's an average speed the difference between the average speed and average velocity total distance in specific time interval it is defined as the average speed or average speed is always positive or sometimes we can have it value as zero average speed is always positive or sometimes we are having the value as zero average speed the or you can say average value ct displacement per unit time now what is an uh difference between speed velocity average speed and average values speed and velocity can be it is not for sure right that speed and velocity are only defined for entire path no it can be in between any two instances while the average velocity and the average speed are defined for the entire path that's why the word average is used net displacement to the total time so if we are talking about the average speed and average velocity it means we are talking in between the two time instances from where the motion started and from where it is going to till uh the point it is going to end not from where okay from where it started until the point it is going to end from where it started until the point it is going to end average speed is a scalar quantity average value ct is a vector quantity average speed of the particle is finite it cannot be zero yes practically it cannot be zero but when the object is stationary right we can't define average speed okay and it is only zero when it is at the rest okay average velocity of the particle can be zero can be positive or it can be negative average speed is greater or equal to the magnitude see i'm again highlighting this statement because on basis of this right sometimes the theoretical questions are asked to do the examination okay average value city is a magnitude and it is always less than the speed of the particle okay so this is a very very very simple explanation in between the average speed and the average velocity instantaneous at what instance the pop the what will be the see at a specific time instant what is the position of the particle at specific time instant what is the position of the particle its specific time instance what is the position blah blah blah lights of those things instantaneous speed and instantaneous velocity gives the value of speed and velocity suppose from t is equal to zero the motion starts at t is equal to t that's the motion ends in between you can take any random instance at that suppose you start from t is equal to zero like from your home if you are having journey to goa obviously it is going to take more than 10 hours right so it is equal to fourth hour what is your average it's what is your speed that is instantaneous speed at t is equal to fifth hour at that moment of time what is the speed what is the velocity that is your instantaneous speed and instantaneous velocity at that moment of time how you are going to define that stuff right so that is the instantaneous speed and instantaneous velocity generally instantaneous speed and instantaneous velocity are defined right and more used in calculus method okay generally they are more used into the calculus method see in the diagram it has been shown right in the diagram it is shown suppose the particle starts as motion from t is equal to zero right at this time instance the speed and the velocity are the instantaneous speed and instantaneous velocity right instantaneous speed is a scalar quantity instantaneous value ct is a vector can be this is positive negative in all of those okay so these are the sum of the motion parameters we discussed now let's move on to the sums right based on it [Music] in one second a particle goes from point a to point b moving into the semi-circle of the radius one meter the magnitude of the average velocity will be what now see the average volatility they are asking right particle is moving on from point a to point b average velocity they are same so average velocity i must define it as right the net displacement by total time in one second it is covering the net displacement of see this is one meter of the radius this is one meter the total displacement is 2r in one second so it is average velocity is 2 meter per second i'm sorry the motion of the particle is defined as this the instantaneous velocity will be what see it's a very simple explanation x is the position bt square the position of the particle changes with respect to this the value the formula of the instantaneous velocity is v instantaneous that is is equal to differentiation of the position with respect to time so can i differentiate this equation with respect to time then i am going to have dx by dt that is v instantaneous so that is is equal to 2b of t and what second at 3 second you have to write at 3 second so we instantaneous right at three second it is is equal to two what is the value of b into three into three three three's on nine twos are eighteen it is eighteen centimeter per second instantaneous word right it is more used right into the calculus method and that entire calculus method is like this way dx by dt and all of those stuff great now directly we are going to move on to the classification of the motion from now onwards we are going to deal with the three kind the uniform motion right the uniformly accelerated motion the non-uniform motion and apart from this in all of those three cases there exists the concept of the relative motion and motion under gravity and this is one right right motion under gravity okay so let's start with the one of the right uniform motion let's starts with the right uniform motion what actual is the uniform motion it's a very simple definition and the entire concept is based on only one formula uniform motion a very simple concept based on single formula right so i must write it down without wasting the time if i want to define the uniform motion then equal distances in equal time intervals equal distances in equal time intervals let's say uh take an example the object starts from a and z b right object starts from a ends at b this is a first time interval t 1 this is t 2 this is t 3 right so from 0 to t 1 the distance covered is d right with the value ctv okay d1 v1 motion parameters may get changed right but the uh the uniform motion states that from 0 to t1 0 to t2 or in another sense from t1 to t2 t2 to t3 in all of those the velocity of the particle is constant right uh instead of value ct let's say speed is constant and it is only based on the one simple formula v is equal to okay speed is equal to distance by time and velocity is equal to displacement by time only one simple formula this is only the simple concept of the uniform motion is all about right okay on basis of this see uh only one question solving approaches there into the uniform motion okay maximum number of the questions are going to come out from the average speed you're going to see okay if i'm talking about the average speed then we are having the uh right definition is total distance covered by total time total distance covered by total time what is the total distance divin d2 d3 suppose the entire path is divided into the n intervals what is the total time then average speed is defined in this way suppose if the distance in the time intervals are not given the value ct and the time intervals are given okay then we have to use this formula v1 t1 plus v2 t2 plus dot dot dot plus vn tn write whole divided by t1 plus t2 plus dot dot dot plus tn suppose if you are not having any uh detail or the data about the time thing okay then you can uh have it like only distance and the value ct intervals are given then we can use this formula it's given by v1 d2 by v2 it's dn by vn so any among this three depending on the whatever the data it is given into the question we can directly use the formula into the specific question apart from this in the case of the uniform motion we are having some of the spatial cases also right what are those special cases first suppose particle starts from a ends at b right it covers the distances into the right two intervals and those two integrals have the equal measurements so we can define it as a case of equal distances suppose here for the first interval the speed is v1 for the second it is v2 then what will be the average speed so the average speed we are going to define it as in this case we are having the direct formula for the equal distance interval that is 2v1 v2 whole divided by v1 plus p2 right this is the formula of average speed second is an equal time intervals distance doesn't matters object starts from a and zb for time t1 it is moving on with speed v1 for time t2 it is moving on with the speed v2 for time t1 it is moving on with the speed v1 for time t2 it is moving on with the speed v2 yes so in this case how we are going to define the direct formula of the average speed right so the average speed it is going to be v 1 plus v 2 whole divided by 2. right so these are some of the important formulas that are used into the uniform portion and the question based on this are asked with an examination the question based on this are asked into an examination okay let's move on to the first question onto it a particles covers half of the total distance with speed v1 and rest half distance with speed v2 the average speed during the entire journey will be what okay now we are going to see the proof of the equal uh distance interval case right i give you all the formula of it now we are going to see the proof suppose it starts from a it ends at b for the first distance d it is moving on with the speed v one for the second distance d it is moving on with the speed v two so what is an actual formula of average speed see only d and v are given so directly we are going to take the formula d1 by d2 whole divided by d1 by v1 plus d2 by v2 right because nothing is mentioned right no data about the time intervals or the time things are given okay so directly we are going to see it's d plus d whole divided by d by v 1 plus d by v right so from here d will get cancelled and the average speed right it is going to come upon as 2v1 v2 right whole divided by v1 plus v2 this will be the perfect answer for this specific question a very simple proof a car moves from x to y with the uniform speed v u and returns to x with the uniform speed v d the average speed for the round trip will be what same it starts from x it reaches to y with the uniform speed v u and it returns with the uniform speed v d it's moving distance into the front direction into the back direction d into the front from x to y and d coming to the back from y to x again it is a case of the equal distance interval so the average speed it is going to come upon as 2 v of u v of d whole divided by v of u plus v of t right d is a perfect answer for this specific question a bus traveling the first one third distance with the speed of 10 the next one third at the 20 and the last one third at the 60 what is average speed now this is a very simple now we are having the three equal distance interval we have seen till now d by three d by three d by three right one with ten second with twenty third with sixty directly we are going to use the formula of the average speed that is is equal to d by 3 d by 3 d by 3 whole divided by d by 3 v 1 d by 3 v 2 plus d by 3 v3 okay so this is going to be 3 whole divided by 1 by 10 how much it is 10 20 and 16 1 by 20 1 by 60. right so final the average speed it is going to be is equal to 3 whole divided by right 20 plus 10. or uh in easy terms we can say it as 1 by 10 plus [Music] this is 60 and this is 3 by 60 right lc of 1 by 20 and 1 by 20 into 3 sorry 4 by 60 right okay 4 by 60 i'm sorry for that right 3 plus 1 4 4 by 60 right so this is gonna be approximately three whole divided by it's 10 by 60 60 into 3 by 10 right this is gonna get cancelled right average speed is going to come up on is 18 kilometer per hour right average speed is going to be 18 kilometers per hour we are having the direct formula for the three equal distance intervals also but rather than to move on to the quite complex approach because right by remembering the formula of the spatial cases also so by the end right you will have oh ho ho ho you'll have that kind of psychology we are having lots of formulas in physics i don't don't want you all right to reach up to that position so for that reason you have to just follow the general concepts right and things will get quite easy a car moves at a distance of 200 meter it covers first off of distance 40 and second half with v now see first half with 40 right second half with v average speed right it is how much 48 kilometer per hour right so again i write it at 48 is equal to 2v into 40 whole divided by right b plus 40 okay so 2 uh 24's are 2 12's are 220's are two tenza two six 6 by 10 is equal to v by v plus 40 6 v plus 240 is equal to 10 v 240 that is equal to 10 v minus 6 v that is 4 v v it is going to come upon us 60 meter per second sorry it is kilometer per hour right it is kilometer per hour b is the perfect answer for this specific question right okay b is the perfect answer from this specific question okay from now right a new thing is going to start that is the uh the uniform motion is quite easy okay it's not a big deal to understand the concept of the uniform motion right things always gets changed and complex right when we are dealing with the concepts of the uniformly accelerated motion and the calculus methods right uniformly accelerated motion and the calculus method okay before moving on to the concept of the uniformly accelerated motion let's understand what actual the acceleration is all about the time rate of change of velocity it is defined as the acceleration okay the time rate of change right in velocity it is defined as acceleration time rate of change of velocity in another sense if if i want to write then acceleration is constant right then i must say that is delta v by delta t is constant velocity is increasing or decreasing at the uniform rate then i can say this as acceleration being constant as the acceleration is generally defined as delta v by delta t change in velocity per unit time we have seen the example suppose object is starting its motion from t is equal to zero right object is starting its motion i'm sorry for this right at t is equal to zero now suddenly what happens let's take an example from a point it start its motion right and it reaches up to point b from a point it starts its motion and reaches up to point b now see in first time interval the velocity is v1 suppose of it started from zero and it reached to do 10 meter per second in time t1 let's say in another time t1 it reached to 20 meter per second in the third time interval t1 it reaches 230 meter per second into the fourth time interval t1 with respect to the reference point a is equal to 0 right it reaches up to the right 40 meter per second so at every on the span of every t1 seconds the velocity is uniformly increasing right mark my words uniformly velocity is uniformly increasing then we can say acceleration is constant apart from this if we want to define average acceleration then it is change in velocity right for entire path by total time change in velocity for entire path by total time instantaneous acceleration acceleration right okay at the specific instance of the time okay acceleration at the specific instance of the time the smaller time rate of change of velocity right at specific instance of time the value of acceleration is instantaneous acceleration the value of the acceleration it is known as what instantaneous acceleration so the general definition of the acceleration is the rate of a change of the velocity with respect to time if this change of a rate is a uniform acceleration is constant it is positive if velocity is increasing it is negative if velocity is decreased average acceleration for entire path i'm talking about let's say it starts from a ends at b for entire path what was the change in velocity with respect to the total time that is average acceleration instantaneous from entire path right i started from t is equal to zero from home after four hours i am going to reach up to m right then what will be the value at m what is the value at surah what is the value of english what is the value at vadra what is the value at anan what is the value at nadia that will define this stuff right that will define this now here comes the right this was a simple definition of the acceleration now here comes the concept of a uniformly accelerated motion the word itself states that object is uniformly accelerated it is uniformly accelerated right or object is uniformly accelerated or how let us see an example object starts from point a it reaches to point b this is time interval t 1 this is time interval t2 this is time interval t3 this is t4 and this is t5 right in between this is my reference point a is equal to 0 from right object started its motion this is my reference point right from t is equal to zero from where the object started its motion this is my reference point so what happens here it's a very simple thing right okay the time rate of change of velocity here it starts from zero it reaches to v now it is going to reach to 2v it is going to reach 3v it is going to reach 4v it is 5b or moving on from lets say suppose object if it starts from a it moves to b we can say it is a case of acceleration why uniform acceleration why because the velocity is uniformly increasing with respect to time but when we say from b to a it is a case of the deceleration or we can say it is a retardation why the velocity is uniformly decreasing with respect to time right it is uniformly decreasing with respect to time so what is the case of the uniformly accelerated or the data motion right in which the acceleration of the system is a constant right okay acceleration of the system is constant right it can be positive it can be negative it can be zero zero is not so possible right it can be positive it can be negative right okay for positive it is known as acceleration right for negative it is known as retardation everything is physics right everything is in physics is described in terms of mathematics everything in physics is described in terms of mathematics everything then how we are going to define this stuff how we are going to mathematically represent this right for that okay we are having two ways the equations right of the uniformly accelerated motion can be described in two ways one the method that is given into the ncrt way using graph equations can be obtained right using graph equations can be obtained second we are having the right the second method right okay the second method it is the first one is an ncrt way right okay the second method what we are having is the calculus method the calculus way first of all i want to tell you all the calculus method is not generally it is not only for uam the uam stands for uniformly accelerated motion the calculus method is not generally for uniformly accelerated motion okay it is for all kind of the motion for all kind of motion right where the motion parameters where the one motion parameters write one motion parameter change depend on another displacement is depending on time velocity depends on time acceleration depends on time like these things are going on like this things right are going on so whenever c calculus method is for all kind of the emotion uniformly non-uniform correctly linear curvy linear 2d everything everywhere we can use the calculus but this is the best way to define best and another way to define the equations right okay generally it is for very small change in parameters right question solving approach for acceleration to be constant right okay if we are directly moving on to the graphical method we are having the first equation v is equal to u plus 80 s is equal to u t plus half a t square right and v square minus u square is equal to 2 a s right so here what is going to happen with respect to the everything is going on with respect to the reference point everything is going on with respect to the reference point if you want to calculate a velocity at any time instance right velocity at any time instance then we can say you can use this equation right displacement with respect to the time uh this reference point at any time instance right or in between the two time intervals right okay and v square minus u square between two in any distance interval s and suppose if we are moving on to the calculus method in cal the entire calculus method right of the kinematics is depending on only three formulas if we are not dealing with the cost then entire calculus method is depending on only three formulas acceleration is dv by dt second velocity is dx by dt right and the third one that is is equal to okay acceleration is dv dv by dx right on basis of this thing the entire calculus method is defined so this is only the question solving approach for all of the kind of the questions right i'm moving aside on the countdown of 10 seconds if you want to just note it down you can 10 9 8 7 6 5 4 3 2 1. particle using those equation we are going to solve that stuff particle starts as motion from rest under the action of constant force constant force it means constant acceleration the distance covered in first 10 second is s1 first 20 second is s2 see it starts from reference point t is equal to 0 in first 10 second it is covering the distance of s1 in next in the total 20 second it is covering the distance of s2 and how we can say what is the relation between the s2 and s2 it starts its motion from rest u is equal to 0 constant force see those equations are only applicable the equations of the uniformly accelerated motion are only applicable in the cases right where the acceleration is constant so here it is mentioned that acceleration is constant directly see this the relation between the time and the displacement it is mentioned so directly we are going to use s is equal to u of t plus half a t square now when this u is 0 because it starts right so now for first 10 second s1 that is is equal to half of a 10 square for next s2 that is is equal to half of a 20 square right so taking ratio can i consider this is equation what can i consider this is equation 2 so s1 by s2 that you are taking 10 into 10 right by 20 into 20 okay so 4 s1 that is is equal to s2 this is the correct relation it is going up the b is the perfect answer for this specific question right the simple use of the equation i'm moving aside simple use right the distance travel by the particle starting from the rest and moving with the acceleration of four by three in the third second will be one starts from t is equal to zero we want to calculate the distance at three second acceleration it is moving it's four by three meter per second and square and it is constant it is a constant it means we can apply the equations of the uniformly accelerated motion it starts from rest it means initial velocity is zero directly we are going to use this equation s is equal to ut plus half 80 square right u is zero s is equal to half acceleration is four by three into nine three is a two twos are s is equal to six meter the total displacement with respect to the reference point is a six meter c is a perfect answer so our sums are going to be quite simple don't get worried whatever the 30 questions are included into the dpv they are wrap check sums to clear out the concept and the basics right it is important car accelerates from the rest at the constant rate alpha right which decelerates at the constant rate b time comes to rest if the total time ellipse is d then the maximum value c d will be what this is a very nice question what it states it starts with t is equal to 0 from rest u is equal to 0. it is getting accelerated by alpha it means velocity from zero is slowly slowly slowly slowly slowly slowly slowly slowly getting increased to max for time t now what happens slowly slowly slowly slowly slowly it is getting right decelerated by beta and after time two again the velocity is getting zero start from rest reaching up to maximum value it will take time t1 and getting accelerated by alpha slowly slowly slowly it gets detailed decelerated by b and the velocity from max it is again coming back to zero so in summarized form it started from rest reach up to the maximum value again it come back to the rest in this case we need to calculate the value of the maximum velocity now what is the total time total time time of the acceleration is t1 time of the deceleration is t2 journey is over the total time is t1 plus t2 and that is t can we use the first equation of the motion here v is equal to u plus 80 for time interval 0 to t 1 can we say final velocity is v max initial value c d is 0 acceleration was alpha times t 1 so can i say t 1 that is is equal to v max right whole divided by alpha can i consider this as equation number one yes i can consider this as equation number one when it returns back first when it was moving forward the velocity was increasing now let's apply the equation again v is equal to u plus 80 now when it returns back initial velocity is v max final velocity is 0 initial value ct is vmix it is getting decelerated beta is a deceleration right alpha is an acceleration acceleration is having the positive sign deceleration is having the negative sign so it is t2 t2 you are going to get it as v max by beta total can i consider this is equation number two total time is t so t is equal to v max write 1 by alpha plus 1 by beta right so v max you are going to come upon as alpha beta of t whole divided by alpha plus beta this is the value of the maximum velocity d is the perfect answer right i'm moving aside on the countdown of 5 seconds 10 seconds you can check it out 10 9 8 7 6 5 4 3 2 1. a body path particle moves into the straight line with a constant acceleration its velocity right changes from 10 to 20 while passing through the distance of 135 in t second value of t is what s is 135 right 10 to 20 v is 20 u is 10 first of all he has use the third equation of motion right to find out the acceleration right it's 400 minus 100 whole divided by 2 into 135 right acceleration it is going to come upon as 300 whole divided by 2 into 135 right so it's 2 15 into tenza 15 9's are 135 right acceleration it is going to come upon as 10 by 9 meter per second square right what is the time t simple again we are using the first equation of motion v is equal to u plus 80 20 minus 10 is equal to 10 by 9 into t right t is equal to 9 second it is coming upon us the answer in the span of nine seconds right this parameters are getting changed body starts from red what is the ratio of the distance travelled in fourth and third second here comes the new formula for the displacement into the nth seconds the displacement into the nth second is given by right it starts from rest u is equal to zero the formula of the displacement of the nth second it is a by two two f n minus one right okay u is zero in fourth and third second into the fourth second it's a by two two into four eight minus one is seven in third second a by two right two into three six minus one is five right so it's fourth by third second so s fourth by s three that is equal to 7 is to 5 a is the perfect answer for this specific question right i'm moving aside see this is how simple they can be right so i'm moving aside on the countdown of 10 seconds you can move particle moves a distance x in a time according to the equation see now here one motion parameter is depending on another then immediately you should all have a click right that the calculus method is going to used particles move distance x is equal to t plus pi whole to the power minus 1 1 whole divided by t plus five x value is this the acceleration of the particle is proportional to see whenever you have to move on from displacement to acceleration why a velocity always do differentiation whenever you have to move on from acceleration to displacement the reverse process why a velocity you have to do integration in integration integration are of the two type definite and indefinite right so whenever you are doing the integration the initial values the constant values are needed okay i think i will uh show you all into the graph based question don't get worried so x is equal to 1 by del x differentiate with respect to time velocity is equal to dx by dt that is is equal to minus 1 t plus 5 whole to the power minus 2 again differentiate this with respect to time to get the acceleration it is dv by dt that is equal to 2 t plus 5 whole to the power minus 3 here acceleration in with respect to time we got but in option it is shown acceleration with respect to distance or velocity we need to mention so what we are going to do right what we are going to do we are going to follow the very simple approach right what will be that simple approach acceleration that is is equal to 2 t plus 5 whole to the power minus 1 multiplied by 3 can i replace velocity here right value ct will be half off right that will be uh minus one it's okay it is directly proportional to t to the power right of minus one replacing it here acceleration is directly proportional to t to the power phi i'm sorry value ct whole to the power 3 by 2 this is the dependency of the motion parameters right this is the dependency of the motion parameters and this is a way right smartly you have to apply the calculus method but don't get worried in the graph conversions i will give you all the brahmastra of the calculus method right into the graph conversion i'm going to give you all the brahmastra of this now see particle moving along the x-axis has the acceleration f and the time is given by this see equation of deceleration it is given as f is equal to write f of 0 1 minus small t by capital t right 1 minus small t by capital t when f 0 and capital t are the constants particle at t is equal to 0 as a zero velocity right what will be the particle velocity in between the time interval t is equal to zero and when f is equal to zero acceleration is given we have to reach up to the velocity then one thing is sure we have to do integration right but in what time interval we have to do the integration in between t is equal to zero and f is equal to zero so first of all what we need to do we need to define the value of t when f is getting zero so when f is getting zero right one minus t by t so small t is going to get equivalent to t so we need to define the velocity in between the time interval t is equal to 0 to t is equal to capital t right so can i say dv by d of t that is equal to f of 0 1 minus small t by capital t so dv is equal to f of 0 right minus f of 0 of t by t right okay into d of t can i integrate this right that is f 0 of d t right limiting from zero to capital t minus f zero of small t by capital t integration zero to capital t dt right zero to capital t dt so here that this is the thing i was telling about whenever you are doing the differentiation no limits are needed but whenever you are uh having the integration limits is needed as what is going to happen right you will have the constant term if limits are if limits are there then this is the case of the definite increase integrals if limits are not there then you are going to get into the mess right so this is going to be f 0 of t minus f 0 of t by 2 the velocity will be f 0 of t by 2 right this will be the perfect answer for this specific question and the perfect use of the calculus method right displacement of the particle depends on the time by the equation see one motion parameter depends on another motion parameter and this the change we are not able to define immediately we are going to use the calculus method the displacement of the particle when velocity is zero displacement of the particle when velocity is zero first of all let's obtain the uh equation by rearranging the term x we are going to get it as t minus t square t square minus six t plus nine this is our dependency of the position with respect to time what we need to calculate we need to calculate the displacement when velocity is 0 first of all let's differentiate with respect to time to get the equation of velocity velocity is 2t minus 6 is equal to 0 at t is equal to 3 second we are getting the v 0 velocity then what will be the value of the displacement at t is equal to 3 seconds the same thing it is asana when velocity is 0 calculate displacement at what time velocity we are getting 0 t is equal to 3 second replace in this equation to get the value of displacement 9 minus 18 plus 9 0. this one question is comfort from my side you can answer beneath the video right okay you can answer beneath the video okay now we are going to move forward or on to the next concept now see uh the graph based questions right for all kind of the emotion and the motion under gravity and the relative motion these things are remaining so we are going to cover between three and three and half hours of this lecture right so let's move on to the next concept of it graph fundamentals so till now in this lecture we have seen uh the fundamentals of the motion into the 1d and then we move forward to the on the motion parameters and then we move forward to the concepts of the uniform motion and then we move to the uniformly accelerated motion now we are directly moving forward to the graph fundamentals now how much part is remaining graph fundamental is remaining and after that we will deal with the motion the gravity cases okay and at the end with the relative motion okay so relative will be the awesome thing now let's move on to the uh graph fundamentals what does it states okay before moving on to that much part okay i'm going to tell you all about okay we need to discuss some stuff right and that even into the detail manner okay first of all uh before moving on to that much technicalities first we will move on to the some of the basic things what is the slope of the graph what is the area of the graph how slopes are defined how areas are defined so first of all if i'm talking about the slope right see suppose here we are having the graph on the y-axis we are having any physical quantity on the x-axis we are having another physical quantity right now the graph is a straight line substanding theta right okay this theta is in anticlockwise direction with respect to the positive x-axis is in anticlockwise direction with respect to the positive x axis right so here in this case how we are going to define the slope slope that is is equal to tan of theta that is is equal to del y by del of x in graph any quantity that is only y with respect to x always gives up the slope this thing you all are aware of right it always gives up the slope right now let's move on to the next thing suppose if the graph is curve then how you are going to able to how you are able to calculate the slope okay suppose this is y versus x graph and it is a curve you want to calculate the slope at the single point x y what you have to do here to draw the tangent if the curve is crossing through single point right then draw the tangent this tangent will give you the idea of this slope right this tangent will give you the idea of the slope so slope here it is going to be d y by dx and that is equivalent to 10 of theta suppose if the graph is crossing through two points right this is x two comma y two this is x one comma y one right the graph the curve is crossing through the two points how you are going to identify the slope in that if it is crossing through one point draw the tangent if it is cross crossing through two points draw the second this is all you've seen into the basic mathematics right okay draw the second right this will give you the idea of the slope okay this will give you the idea of the slope okay so when crossing through the one point draw the tangent and crossing through the two points draw the second okay so here we are going to have it as y2 minus y1 whole divided by x2 minus x1 that is equivalent to tan of theta and that is is equal to slope this is how things are going to get calculated this is how so slope is defined in each and every graph if it is a curve right if it is a parabola kind of thing directly you can move on right if it is not in parabola an irregular curve crossing through more than one points then you can draw the seconds right okay and uh lastly for the straight line you can calculate this slope right so this was the first fundamental of the graph that is known as slope let's move on to the next one okay question solving approach it is one and the same thing but first of all we are going to see write the things into the graph and we are going to see into the graph conversions okay now how you are going to calculate the area for this symmetric and the asymmetric graphs suppose this is a graph one physical quantity on the y-axis the second physical quantity is on the x-axis now what is is going to happen suppose the graph is like this way right then what will be the area under the graph area under the graph will be this much area that has been right shaded area shaded under the graph that will be equivalent to right the area of the symmetric shapes here rectangle may appear right triangle may appear right anything may appear so area and that will be obviously equivalent to y del x right that will be obviously equivalent to y del x if it is of the symmetric shape how to calculate the area for the asymmetric shapes in between the two limits right how to calculate the area into the asymmetric shape let us take another example of the graph right this is the y-axis this is the x-axis suppose the graph is like this way okay the nature of the graph is kind of the parabolic or a kind of a curve generally parabola is used to write take a as an example why because students used to get right easily used to get with it right easily used to get comfortable with it so that's why the parabola is used but in this case okay you can consider it as a curve also right so here how we are going to calculate the area first of all define the two limits suppose this is an initial limit right okay this is a final limit right so through the drawing of the tangent by slope we will get relation between y and x by slope we will get the relation between y and x so what happens here right area that relation we are going to replace in this equation for the curve and what will be the limits limits will be from x of initial to x of final right so by this equation how much amount of the area we are getting okay the region that is shaded here okay so this is how you are going to calculate the area into the symmetric shapes right and this is how you are going to calculate the area into the curves okay so these are the two most important fundamentals of the graph okay this are the two most important fundamentals of the graph now we are going to see okay what are the graphs of the uniform motion what are the graphs of the uniformly accelerated motion what are the graphs of the non-uniform motion okay all of those things we are going to see and after that we are going to deal with all of those sums right okay before moving on to the graph conversions let's see all of those graphs first the graphs of uniform motion graphs of uniform motion okay what does it states uniform motion means it is a very simple thing velocity of the particle is constant or you can say speed is caused okay so it means that its position versus time graph is going to be like this way right in which this is a slope that slope is constant the velocity versus time graph is like to be this way because it is a uniform motion constant now this is a rectilinear motion so we can say the speed and velocity both are constant generally if we are talking about the motion in the 2d right then we can't say the velocity as a constant we have to say speed is a constant but here as we are dealing with the motion on the rectilinear path right we can say the speed and velocities are constant because the direction of the velocity is not going to change in that okay if it is going to change then the graph is going to appear on the negative side what will be the acceleration versus time graph right this will be zero acceleration versus time graph will be zero yes or no right okay now second will be the graph of uniformly accelerated motion uniformly accelerated motion states that right acceleration of the particle is kept constant right acceleration of the particle is kept completely constant so here on the right side i'm going to write it as acceleration constant what will be position versus time graph let's say particle starts from rest right if it starts from rest can i write it at x is equal to u of t plus half of a t square suppose it starts from rest can i say x is equal to half of a t square can i name it one can i name it two right okay suppose see first of all if we are taking it as a rest then we are able to understand the actual nature of the graph right the actual nature of the graph x is equivalent to half a t square acceleration is constant it means x is directly proportional to t square we are having this kind of the parabolic right this thing you have seen into the basic mathematics x is directly proportional to t square right x is directly proportional to t square okay now suppose this is a second curve suppose if you want to plot it up x is equal to u t plus half a t square in which u t will be always ut will act like an intercept that x into the straight line right okay will be the fixed value right then the graph is going to appear like this way this right okay i'm sorry uh graph appears to be like just hold on okay i'm sorry for that okay graph appears to be like this way this will be u of t this is a first graph acceleration is a constant and depending on the three equations that we obtain into the cases of the uniformly accelerated motion we plotted up the graph okay now the second equation is of v is equal to u plus 80 i am naming it as a one second suppose u is zero then v is equal to eighty v is equal to u plus 80 right it means we are having u as the intercept right we are having u as the intercept so directly we can say the first graph would appear like this way okay this will be the u intercept and this is how the velocity is increasing this is a fix right beyond this the velocity is increasing with respect to time okay what about v is equal to 80 this will be the first graph this will be the graph of v is equal to 80 if it starts from rest right if it starts from rest so this will be the second graph okay so this is how the graph for the velocity versus time into the uniformly accelerating if velocity is increasing with respect to time then graph is a straight line if it is decreasing with respect to time then we have making the negative slope here we are having the positive slope right here we are having the positive slope but when the velocity decreases with respect to time we are having the negative slope velocity decreases with respect to time we are having the negative slope lastly what will be the acceleration versus time graph the third graph acceleration is constant right so either it can be positive that is acceleration greater than zero or we are having the negative acceleration that is acceleration less than zero but the value is going to get constant the value is going to be constant right so these are the graphs of the uniform motion right the next one it is a graph of the uniformly accelerated motion i'm moving aside on the countdown of 10 seconds right if you want to just note this down you can note it up right 10 9 8 seven six five four three two one that's it okay now the basics of the graph we saw right the fundamentals of the how to calculate the slope how to calculate the area right we saw okay after that we deal with uh this thing what we can say uh the types of the graphs now we are actually moving forward to the graph conversions right we are actually moving forward to the graph conversion now what does this state what is a master of graph conversion first of all we are going to deal with the entire theories of the graph right in some of the resources thousands of books are having plenties of right graphs xt speed time position time like this and that and okay unnecessary list of graphs are given i'm not saying the content is not useful right content is useful but basically it is creating a high impact on your mind okay basically it is creating the high impact on your mind okay so that differs that thing differs okay that thing differs right so yeah graph what is the brahmastra of the graph conversion the brahmastra of the graph conversion is fundamentals of graph plus calculus method right fundamentals of the graph plus calculus method this is a brahmastra of the graph right how in calculus method we are having only uh three simple formulas one that is a is equal to b dv by dx right second is equal to a is dv by dt right and v is equal to dx by dt fundamentals of the graph right always see in the sums of the i'm going to show you all the sum sums of the graph conversion okay how i am saying this right and why i'm saying this it is important okay so for that let's solve some of one example suppose right [Music] the v versus x graph is given v versus x graph is given the question states that okay make the appropriate a versus x graph in some of the questions right as we are dealing with mcq so they might be having uh the four options also they might be given right that might be given the four options okay but what how to deal how to reach up to the exit answer that is important okay see immediately we can say this is a graph with a straight line okay so the step one is identify the nature of the graph whether it is a straight line it is a curve right what kind of the curve it is a parabola hyperbola ellipse right okay this all thing it is a circle all thing you all are aware of okay what is the equation of the circle what is an equation of ellipse okay what is an equation of the parabola kind of parabola hyperbolas okay the equation of the straight line you all are aware of so first step is identify the nature of the graph question solving approach i am writing here okay first step identify the nature of the graph what will be the second step obtain the relevant equation based on the nature obtain the relevant equation from it this is a graph of the straight line okay it is having the intercept so it means immediately right it should be clicked then we are going to follow the equation y is equal to mx plus c on y-axis we are having physical quantity velocity on x-axis we are having the physical quantity uh this x intercept is v0 what about the slope okay you let's change one value let's have intercept as u this will be v 0 and this will be x 0 right okay so what is the slope of this graph right the high is v 0 minus u whole divided by x 0. so here we are going to get it as v 0 minus u whole divided by x 0 so according to y is equal to m x plus c we obtain this equation again i'm rewriting v is equal to v 0 minus u whole divided by x 0 right if you don't want to make that much complex at the initial level right okay you don't want to make things right then you can remove the intercept right you can remove the intercept if you are removing the intercept things are going to be quite easy suppose the graph starts like this way okay then y is equal to mx plus c right obtain the relevant equations of it okay why is we slope is positive v 0 by x 0 right okay we are having x plus no intercept okay no intercept right can i consider this equation number one so step two successfully completed step one was identify the nature of the graph step to obtain the relevant equation from its nature okay step three use the calculus method use the calculus method now suppose we obtain the relation right v versus x right so this is a given data now what we have to find out try to identify how to reach from v to a via x using the calculus method right so we are having only the three formulas of the calculus method a is equal to v dv by dx right a is equal to dv by dt and v is equal to dx by dt anyhow see if you have to move from displacement to velocity velocity to the acceleration anyhow you have to do differentiation if you want to follow the reverse phenomena then you have to follow the integration so one thing is fixed we have to move from velocity to acceleration it means we have to follow the differentiation but it is important to select the correct formula from the calculus method so we have to move from v to a y i x this is only the method that can be useful sorry this is only the formula that can be useful let's see okay so let's differentiate with respect to x okay so it is dv by dx that is is equal to v0 by x0 right this is not acceleration consider this is equation number two acceleration is a is equal to v dv by dx replacing the formula of v replacing the formula of dv by dx so you are going to get v 0 by x 0 into x into what is d v 0 by x 0 so acceleration equation is v 0 square by x 0 square into x here v 0 and x 0 are constants v 0 and x 0's are kept constant yes or no do you got this stuff everyone those who are like do you got this stuff on now step four we successfully completed the last step of the graph conversion right okay this is the fourth on basis of equation we obtained from calculus method select the correct graph select the correct graph it means it is very simple thing we obtain the equation just mark my words right equation is a is equal to 0 square by x 0 square into x this is a constant term can i say a is directly proportional to x it means it is a graph of this straight line with 0 intercept just the slopes are getting different this is a versus x right this is a graph right this is a theta here the slope that will be equivalent to v 0 square by x x0 square right and this is the a versus x crop with the zero intercept right okay here the slope is b0 by x0 this is only the straight line graph here also we are getting the straight line but the slope is v 0 square by x 0 square there is a difference of slope in between these two okay this is the actual phenomenon of the graph conversion okay so students uh first of all let's summarize this thing into the let's summarize this thing right step one identify the nature of the graph step two obtain the equation based on the specific nature step three apply the relevant calculus method step four obtain the relation right this is a relation where we have to reach and that relation that has been asked in a question right after reaching up the relation identified whether it is the relation matches to the equation of the straight line whether this relation matches to the equation of the ellipse where this relation matches up to the equation of circle right then draw the appropriate graph of it this is a very simple phenomena of the graph conversion right a very simple and appropriate phenomena of the graph conversion yes or no okay now let's see some of the sums based on it okay some of these sums we are going to see some seven to eight sums right on the brace of the graphs and the graph conversions right on the basis of the graph and the graph conversions right so are you all ready the particle shows the distance time curve as given into the figure the maximum instantaneous velocity of the particle around the point will be what maximum instantaneous velocity see first of all i'm giving you all a 10 second do like watch just stuff to pause this video right and try to it's 10 9 8 7 6 5 4 3 2 1. particles show the distance versus time graph and we see from distance anyhow we have to reach to the velocity it means anyhow we have to deal with the slopes because any our ratio way of distance to the time is going to reach help us in reaching the i'm sorry instantaneous velocity suppose if i'm talking about point a right we want that point among a b c d right where the instantaneous velocity is maximum i told you all into the fundamentals of the graph graph is crossing through the multiple points but we have to analyze each point individually graph is crossing through multiple joins right we can select any of 2 and draw the second but the question states which point among four has maximum instantaneous velocity so it means that each and every point should be analyzed individually so for that we have to draw the tangent suppose if i'm talking about the b right draw the tangent this is the angle theta b is making if i'm talking about this c right and if i'm talking about d where you're going to get okay the maximum intensity the point where the slope is maximum right we have dx by dt that is is equal to maximum when dx by dt is equal to maximum we are having instantaneous value ct to be maximum dx by dt to the maximum we are having instantaneous velocity to be maximum dx by dt to the maximum we are having instantaneous velocity to be maximum yes or no okay so here you can understand as compared to b and d the theta c is right the angle theta c is maximum so on basis of the value of the theta c we are having 10 theta c value as the maximum okay so c is the perfect answer for this specific question where we are having the maximum value of the instantaneous velocity right maximum value of the instantaneous value city so let's move on to the next question onto it the displacement time graph of the moving particle is the instantaneously again the displacement time graph came now this time instantaneous velocity of the particle is negative at it is displacement versus time graph immediately it should get click into the mind where it's the slope of the displacement versus time graph is an acceleration sorry velocity so wherever we are going to get the negative slope that point is our answer okay so four points are given obviously right we have to analyze each and every point individually so we are going to drain up the tangent here at c the theta is acute if theta is acute then the tan of theta is going to be the positive value theta is acute then tan of theta is going to be the positive value right 10 of theta is going to be the positive value d theta is 0 slope is 0 instantaneous velocity is zero f theta is acute tan theta is a positive instantaneous value ct is positive right e this theta angle is always calculated in anticlockwise direction with respect to positive x-axis okay anti-clockwise direction with respect to the positive x-axis it means that with respect to positive x-axis this i'm going to theta in the anti-clockwise direction then at position e right theta is obtuse if theta is obtuse then tan theta value will be negative if then theta value will be negative then slope will be negative if slope will be negative then instantaneous velocity will be negative right slope will be negative instantaneous velocity will be negative okay so here uh the e is gonna be perfect answer for this specific question e is gonna be perfect answer for this specific question let's move on to the next question onto it the variation of the velocity versus the velocity of the particle with time moving along the straight line is illustrated in graph what is a total distance traveled by the particle in first four second velocity versus time graph area under this gives distance displacement slope under any symmetric curve gives acceleration this is vt graph right in vt graph it is a very simple thing area under graph will give distance area under graph will give distance right yes okay so how to calculate this is area a1 this is area a2 this is area a3 this is area a4 and this is area 5. so the total area under the graph in between 0 to 4 seconds total area under the graph in between 0 to 4 seconds will give right the actual value of the the distance of the particle so can i say distance between 0 to 4 second is equal to total area right so total area is a1 plus a2 plus a3 plus a4 plus a5 a1 is triangle so it is half base into altitude so it's half 1 into altitude of 20 plus a2 is 1 into 20 plus a3 is half into 1 into 10 plus a4 is 1 into 10 plus a5 is 1 into 10 so this is going to be 10 plus 20 plus 5 plus 10 plus 10 right so this is going to be uh 20 50 55 meter the total distance traveled is 55 meter and b is the perfect answer for this specific question total distance traveled is 55 meter and the b is the perfect answer for this specific question yes or no graph conversion the v x graph of a particle moving in the straight line is shown below which of the below graph shows a versus x see we have seen with the positive slope here comes uh the thing with the negative slope v versus x graph is given we need to select the correct a versus x graph step one identify the nature of the graph straight line step two obtain the relevant equation according to the formula y is equal to m x plus c on y axis we are having velocity slope theta is obtuse right slope will be negative so it is negative of v 0 by x 0 on x axis we are having x plus this is the intercept v 0 second step successfully completed the first step you identify the nature of the graph that is a straight line second step we obtain the relevant equation now third step we have to use the relevant calculus method equation is x naught x plus v naught we have to reach from velocity to acceleration via displacement so it means anyhow write a okay we have to use the formula a is equal to v dv by dx from the calculus concept let's differentiate this equation with respect to x so it's dv by dx that is is equal to negative of v naught by x naught can i consider this as equation number one can i consider this as equation number two right from this we are going to obtain the equation of the acceleration acceleration is equal to v dv by dx right so that is equivalent to negative of v naught by x naught of x plus v naught right into right negative of v naught by x naught so resolving this you are going to get v naught square by x naught square of x negative of v naught square by x dot if you are analyzing this equation anyhow a is equal to mx minus c is there relation of acceleration and your displacement axis is having only one power it means it is going to come the equation of the straight line right only either like the slope may differ or any kind of the uh that will be the change into the values of the intercept so this is going to follow the y is equal to mx plus c right this is going to follow the y is equal to mx plus c right but here we are having the positive slope intercept is negative negative intercept v naught square by x naught slope is positive see with respect to the positive x-axis theta is acute if theta is acute slope is positive this will be the perfect a versus x graph so what we did step one identify the nature of the graph step one obtain the relevant equation step three try to find out which formula from the calculus method we are going to use step 4 apply the relevant or write formula and get the equation step 5 from the equation identify what will be the nature of the graph that has been asking question what will be the nature of the graph that has been asked into this question yes or no d is the perfect answer for this specific question right d is the perfect answer for this specific question is it cool let's move on to the next question onto it a ball is dropped the displacement versus time graph is shown right okay this is uh right the thing that is going to come up into the work energy and power also so we are leaving it here only okay so this was or you can try it into this homework this is the second homework based question i'm going to give up you all the first was given yesterday okay this is a humble from my side okay this is the homework from my site okay we can try or else we are going to see this question into the work energy and power also now let's move on to the motion under gravity case okay motion under gravity is quite simple when object is right thrown or released from some okay what does this galileo stated the galileo stated that when two objects irrespective of the masses released from top of leaning tower of pisa they are accelerated by constant value towards the center of earth the motion of the particle as because of this kind of the constant acceleration is defined as motion under gravity motion of the particle as because of this kind of the acceleration it is known as motion under gravity because this acceleration is gravitational acceleration is because of the earth whose value is g is equal to 9.8 okay whose value is g is equal to nine point 9.8 meter per second square yes or no 9.8 meter per second square yes or no in motion under gravity what will be the question solving approach see only two things are going to be useful into the case of the motion under gravity the first thing that will be um this one what we can say sign conventions and the second one in question solving approach like for this right is the appropriate use of the equation first of all in the case of the motion under gravity i'm going to write some noteworthy points the acceleration is always negative air resistance is completely avoided there are some cases in which the air resistance we are going to write uh like it will be taken into the consideration but right now for maximum number of the cases let's avoid this error resistance okay let's avoid this as resistance so our resistance is completely neglected a is equal to negative g right object is thrown always it is thrown less than the escape speed of the earth it is always thrown less than the escape speed of the earth always right it is always thrown less than the escape speed of the earth right and apart from this if it is thrown less than okay the curvature effect of earth is always avoided for maximum number of the cases as we are dealing the motion and the straight line right for kinematics a will be of negative g the value of g is not changing right the value of the g never changes okay now this was the first most important part regarding the motion under gravity okay the next most important point that we are going to deal with these sign conventions say for sine conventions i had already told you all the vector is acting into the downward direction right it we are taking it as a negative j it will be negative if the vector is acting into the upward direction we are taking it as a positive j it is positive if the vector is directing towards the positive x stick then we have a positive i which is directing towards the negative x axis we are having the negative i okay so these are the same kind of sign conventions we are going to use and i told you all into the right the previous part of this session also the sign conventions are always applied with respect to the reference point and into the cases of the motion into the straight line t is equal to 0 will be our reference point right t is equal to 0 will be our reference point okay t is equal to 0 will be our reference point now in case of the motion under gravity a is equal to negative g is there why because gravitational acceleration all always acts into the downward direction in some of the right resources they mention that downward is taken as positive upward is taken is negative all of those relevant equations are going to form don't get worried but but it is important right to follow the approach that we are going to use into the entire syllabus it is important to follow the approach so for that reason always i am taking downward as the negative and the upward is positive a is equal to negative g obviously it is going to be the constant value right it is going to be the constant value yes it is going to be the constant value so it means that we can apply the equation of the uniformly accelerated motion v is equal to u minus gt second s is equal to u of t minus half of g t square third v square minus u square is equal to negative of 2 gs we have to apply the sign conventions i have only applied now first of all let's see right how sign conventions are applied then and then we are going to see towards the sum right first how to apply these sign conventions into the multiple cases and how to write down the equation suppose object is thrown into the upward direction from the graph object is thrown into the upward direction from the ground right so here you slowly slowly slowly it will move forward it will reach to the max side and it is going to come it follows this straight line but for the understanding purpose we are drawing like this right so how to write down the first equation of the motion see first equation of the motion states v v is equal to u plus a t this will be our reference point t is equal to zero with respect to this we are going to apply v will be the final velocity at the top zero initial volatility is directing into the upward direction positive acceleration is g that x into the diagonal direction it is negative g second equation s is equal to u of t plus half of a t square with respect to the reference point see always whenever you apply the sign conventions now always eyes apply the sign conventions right in between any two points generally it is not even with respect to the reference point i am writing now apply sign conventions any two points any two points you can apply the sign convention right now i've taken the ground and the top is the first point slowly slowly i'm going to change up the values don't get worried so here s will be positive u will be positive g will be negative this is the second equation of motion now suppose if i am taking the second case right from the cliff from a tower object is released okay so this will be the first point this will be the second point between a and b we are going to apply these n conventions right between a and b we are going to apply these sign conventions so what is going to happen here right v is equal to u minus g t sorry u plus g d g will be negative okay u is 0 because it is dropped v will be negative because v is acting in this direction so v is equal to negative of g so the first equation is going to come upon as v is equal to gt the second equation is s is equal to ut plus half of a t square s is negative because it is directing downward u is 0 plus negative of half g t square so s is equal to half of g t square this will be the second equation of the motion how to write down the third equation in between initial and final points right v square minus u square is equal to 2 a s final is v square initial is 0 to negative of g negative of s v square is equal to 2 g s this will be the third equation of motion this is how sign conventions are applied in between two points right this is how these sign conventions are applied in between the two points yes or no right now let's move on to the next question let's see some of these sums right and see right and we will analyze this things right let's see some of the sums and analyze this stuff now a boy standing at the top of the tower of 20 meter height ground tower height 20 meter drops a stone assuming g is equal to 10 it drops a stone it means initial velocity is zero this is a point a this is a point b in between the top of the tower and the ground we are going to apply the sign conventions and get up to the results the velocity with which hits the ground see no data about time is given height is given we have to calculate the velocity and the initial values it is zero immediately it should that that we should get a click that we have to use the third equation of motion v square plus v square minus u square is equal to 2 a s the third equation of the motion that we have to use apply these sign conventions in between a and b u is 0. v is actually negative but we are having square so it is going to be positive j is negative and with respect to a displacement is also acting directing the in downward direction so it is negative so it is 2 negative of g negative of s right so v square it is is equal to 2 g of s or you can say h 2 let's say g is 10 into 20 so v square it is going to be 400 with 20 meter per second velocity it is going to hit up the ground with 20 meter per second velocity it is going to hit up the ground yes or no okay so b is the perfect answer for this specification a ball is thrown vertically upward from the ground it has a speed of 10 meter per second when it reaches one half of its maximum height how high does the ball rise this is a very nice question first of all i'm giving you all the 10 seconds right to uh solve it up right okay if you want to 20 seconds you can you can just pause this video and move on okay 10 9 8 7 6 5 4 3 2 1 you might have paused this video let's give you all another 10 seconds 0 minus 1 minus 2 minus 3 minus 4 minus 5 minus 6 minus 7 minus eight minus nine minus ten okay so hope so you might have paused this video right and a ball is thrown say from ground it is thrown upward with the speed of 10 meter per second it means the initial velocity it is 10 meter per second when it has reached one half of its maximum height suppose here this will be the h max we are talking about the one half of the maximum height it is h max by two okay i'm not i'm sorry uh it is not thrown with the 10 meter per second right when it reaches it is thrown with you when it reaches the one half of the maximum height right it is 10 meter per second how does right how high does the ball rise it means we have to calculate the max height so it is thrown with you at half of the mag side we are having 10 at the max height we are having the velocity as zero so let's apply the third equation of motion in between a and b and in between a and c first let's apply the third equation of the motion in between a and b v square minus u square is equal to 2 a s v square is 10 100 minus u square we are not aware of that will be equivalent to [Music] right uh okay rather than to follow that much long approach right we can directly use the equation we can directly apply to b and c also no because here values it is 10 okay we do not want to calculate the initial velocity we have to calculate the max side right then we are going to see we are going to apply between b and c v square minus u square is equal to 2 a s right for b c this will be the initial value minus 100 at this back side the final velocity is zero 2 g into s s is h max by 2 right h max by 2 height is covering this is a h mix half is a this is half of h max so this one is the remaining half so 2 2 is gonna get cancelled right h max you are going to get it as 10 of meter right simple a is the perfect answer for this specific question a body drop from the top of the tower fall through 40 meter in last two seconds in last two seconds the total time of fall is t in last two second t minus two seconds right it cover ups the distance of 40 meter the height of the tower will be what in last two seconds it is covering up the distance of 40 meter can i say height of the tower is h right can i say height of the tower is h yes it is dropped it means u is equal to zero so can i say h that is is equal to half of g t square right h that is equivalent to half of g t square can i consider this as equation number one if this is h this is 40 then this will be h minus 40 right okay h minus 40 right this will be last two seconds so this will be t minus 2 seconds so can i say h minus 40 that is is equal to half of g of t minus 2 square equation number 2 right equation number 2 right so this is equation number 1 this is equation number 2 right can i replace 1 into so you are going to get half of gt square minus 40 is equal to half of g t minus 2 whole square from this you are going to get the value of t if we are going to get the value of t immediately we can reach up to the h right so this is is equal to half of g t square minus 40 that is is equal to half of g t square minus 2 t plus 4. right so it is half of g t square minus 40 is equal to half of g t square minus half of g into 2 t plus right g by 2 into 4 so this is gonna get cancelled minus 40 is equal to this 2 is going to get cancelled minus g of t this 2 right plus 2 of g right minus 4 g minus 2 g is equal to minus gt g is gonna get cancelled t you are going to get it as six seconds right you are going to get it as six seconds we sold it up yes we solved it so the total time of the journey is six seconds h is equal to the total time of the journey is six seconds h is equal to ut plus half 80 square h we need to calculate u is 0 half into g into 6 square h is equal to a half into g into 36 to 18z height is 180 meters the height of the tower right the total uh time we got it as these six seconds right total time we got it as six small mistake we did right okay here we are going to get it as 40 minus 4 right okay 40 minus 4 so solving this you are going to get t is equal to 3 seconds half g t square minus 40 plus four i'm sorry for that so t is equal to three seconds you are going to get and here right six square is not going to happen it is going to happen as rate happens it happens right it happens human error okay so it's 3 square so h you are going to get it as 45 meter b is the perfect answer for this specific question b is the perfect answer for this specific question right let's move on to the next question on to it what will be the ratio of the distance traveled by the freely falling body what will be the ratio of the distance traveled by in the fourth and fifth seconds of the journey fourth and fifth second of the region directly we are going to use up right the displacement for n second formula right so it will be s of 4 by s of 5 it is dropped u plus g by 2 2 into 4 minus 1 u plus g by 2 into 2 of phi minus 1 u is gonna be 0 freely falling body from rest g by two g by two will get cancelled four twos eight minus one is seven phi two's at ten minus one is nine b is the perfect answer for this specific question right displacement of the nh seconds formula the water drop falls at the regular intervals of the time this is a very nice question i'm going to give you all into the homework right you can try it by yourself okay because uh now approximately uh what we can say more than right uh two hour and 45 minutes of the class has been gone and only this thing is remaining what we can say the relative motion so directly we're going to concentrate on it okay directly we're going to concentrate on it let's uh move on to the concept of the relative motion right entire concept of relative motion depends on only single statement that we have discussed into the right that we have discussed into the beginning of this session right we have discussed what it states nothing in universe is at absolute rest nothing in universe is in absolute motion nothing in the universe is at absolute rest nothing in the universe is in absolute motion everything depends on the frame of reference and how observer sees this stuff entire relative motion depends on this thing entire relative motion depends on this thing yes or no nothing in this universe is at the absolute rest nothing in the universe is in absolute motion nothing in the universe is at absolute rest nothing in the universe is in absolute motion so what does the relative motion states motion of an object with respect to multiple observers is the relative motion right now we are going to see this relative motion in one day into the very simplified form right again it is going to come back into the next session about the relative 2d again it is going to come back right so what does this relative motion states right how the positions of the observers are getting changed how things are getting changed how equations are getting changed how we are analyzing the motion through the difference perspective let's take an example to understand this stuff right suppose there are two cars this is car a this is car b this is moving with 10 meter per second this is also moving with 10 meter per second right okay both are moving with the 10 meter per second right both are moving with the 10 meter per second let's analyze this thing let's change this value 15 meter per second and 10 meter per second i am taking three observers first observer is on the ground observer is on the ground second observer is on a third observer is on b first we are taking observer in ground i am standing on the ground i can see a is moving with some velocity b is moving so velocity yes everything is fine i am on the ground i am able to see the reality the person who is on the ground always sees the reality right so the observer who is on the ground sees the reality so what is the velocity of a with respect to ground this abbreviation means that velocity of a with respect to ground can i say that this is equal to 15 meter per second what is velocity of b with respect to ground 10 meter per second this is a very simple thing the all the times we have seen this stuff let's take observer on a now observer is on a what is the velocity of a with respect to a zero why observer is on the a car a is moving with 15 meter per second observer is also moving into the 15 meter per second right relatively they appear stationary with respect to each other velocity of a with respect to observer is 0. what is value ct of b with respect to a is vb minus va that is is equal to 10 minus 15 negative of 5 meter per second what this negative sign indicates right what does this negative sign indicates see this is the difference when the observer is in the ground they both are moving a is moving and b is moving with different velocities but when we are on the a observer is on the a the speed of the b is less with respect to it so what it happens b is coming nearer the negative sign indicates b is coming backward this appears when observer is on the a it does not appears when the observer is on the ground things are different let's check what happens when observer is on b first observer is on the a i can see i'm having more velocity as compared to b here b you are coming nearer now let's jump off observer is on b right now what will be value ct of b with respect to b zero why observer is on the b car is having the velocity 10 meter per second observer is also having the velocity 10 meter per second and that also into the same direction it means relative zero so they both appear rest with respect to each other now observer is on the v hello a i am coming nearer to you how c velocity of a with respect to b v a minus vb is equal to positive phi right okay okay b is here a is coming nearer oh come come so this positive sign indicates a is coming this side according to the sign convention this side we have taken is positive this side we have taken as a negative so here the negative sign indicates when observer was on a right the negative sign indicates that b is coming backward it appears that see if observer is on a then it appears that b is coming backward if observer is on b it appears a is coming in front like that on the ground what happens both are moving so that is the difference of the right uh the position of an observer that's why it has been told nothing in universe is at absolute rest nothing in the universe is an absolute motion okay nothing in the universe is at absolute rest and nothing in the universe is in absolute motion everything depends on the frame of references okay everything depends on the frame of references right what is the question solving approach see in the concept of the relative motion always two interacting particles are there always two interacting particles are there always right so this is a step one what is the question solving approach for all kind of the check it out anyhow acceleration 1 with respect to 0 acceleration 2 with respect to 0 this is there or not relatively if we are having no acceleration it means it is the case of relative uniform motion relatively none among two is getting accelerated relatively none among two is getting accelerated then it is a case of relative uniform motion in which only one equation we are going to use that we were using into the uniform motion but this time from the observer's perspective we are going to use it from the observer's perspective we are going to use this up any two interacting particles are there v 1 2 is equal to s 1 2 by t suppose if a 1 2 is not equal to 0 a 2 with respect to 1 is not equal to 0 or a 1 2 is equal to constant right or a 2 1 is equal to constant then this is the case of if any one of the interacting particles are having the acceleration then this is going to be non-zero then this will be the case of relative uniformly accelerated motion so in that case a 1 2 or a 2 1 is equal to constant v 1 2 u 1 2 plus a 1 2 into t s 1 2 is equal to u 1 2 into t plus half of a 1 2 t square v 1 2 square minus u 1 2 square is equal to 2 1 2 s 1 2 this is how the three equations of the motions are written defined right this is see this is the same equation we have used but this time in the relative motion we are using from the observer's perspective right in the relative motion we are seeing from the observer's perspective so in summarize form if we are talking about what actual uh the thing is the actual thing is identify whether uh the both of the interacting particles are having vessel see if both of the particles are having the same acceleration then on then this velocity this relative acceleration is going to be zero and it will be the case of relative uniform motion you have to just identify with relate whether the relative acceleration is there or not and apply the relevant equation pre-t uh okay this is a simple thing okay a bus is moving with the speed of 10 meter per second on the straight road bus it is moving with the speed of 10 meter per second on the straight road scooterist he wished to overtake he is distance d behind the bus he want to overtake the bus into the 100 seconds okay that distance it is given as one kilometer with what's paid the scooters should chase the bus i want to ask you all right with what speed right this what will be the speed of the scooterist so that he can cross distance of one kilometer in 100 seconds relatively first of all let's keep the observer on bus right check it out any of them is having the acceleration no a1 is also 0 a 2 is also 0 it means a 1 2 is 0 or a 2 1 is equal to 0 it means it is a case of relative uniform motion case of relative uniform motion so what is going to happen in case of the relative uniform motion right what is going to happen in case of the relative uniform motion right simple equation we are going to see now observer is on this is 1 this is 2 observer is on 2 so velocity of 1 with respect to 2 is equal to displacement of 1 with respect to 2 whole divided by time v 1 of two relatively they have to cross one kilometer it means thousand meter in what time 100 seconds so this is going to be 10. so velocity of 1 with respect to 2 it means velocity of scooterist with respect to bus we are going to get it as 10 okay in this you have to apply the sign conventions i have taken everything is positive why because distance is covering on the right side positive velocity is on the right side both are having so it is positive so never forget to apply the sign conventions that we did into the motion inter gravity so vsb is 10 so vs minus vb is equal to 10 so vs is equal to 10 plus vb so what is the velocity of scooter is 20 meter per second this is how things are getting resolved okay i have given you all the brahmastra of the right why it is taking the less amount of the time for the discussion because i have given you the master of the relative motion train of 150 meter length is going towards the north direction train length is 150 meter it is going toward the north direction at the speed of this is the speed of train that is is equal to 10 meter per second parrot flies at the speed of phi in the south opposite to it parrot is flying this is velocity of parrot five meter per second opposite to it on the parrot the time taken by parrot to cross the road any of them are getting accelerated no acceleration of crane 0 acceleration of pan is apparent is 0 so it is a case of the relative uniform motion relative uniform motion it is a case of the relative uniform motion v one with respect to two s one with respect to two time now see they both are moving into the opposite direction so one velocity will be considered as the positive the velocity of the train and velocity of the parrot will consider as the negative right so here velocity of train minus velocity of parrot that will be equivalent to the total distance is 150 meter into time t now this is considered as positive 10 negative negative of 5 is 150 by t t is going to be 10 seconds in 10 seconds right parrot will cover up the bus the train the t is the perfect answer for this specific question this is how the equations of the relative motions have been solved don't worry the same thing is going to repeat into the 2d and we are going to analyze the concept of bundy again into the next session but this was the thing into the very summarized form for you okay two trains each are 50 meter long they both are moving into the opposite direction 1 into 10 meter per second second is moving with 15 meter per second the time to cross each other check it out rate the relative acceleration of the both is zero it means it is a case of relative uniform motion v one with respect to two s one with respect to two time s when one okay both are moving into the opposite direction right so value cts will be added 25 distance see 50 this 50 this 50 50 hundred first they both need to cross each other first they will get overlapped they will cover the distance of 50. after that to cross each other they will cover up the distance of 50 because both are having the length of 50 meter this will be 100 answer will be 4 seconds b is the perfect answer for this specific question the one this is homework from my side right okay don't get worried things are going to come up into the next session also right okay so students uh it's time to go yes it's time to depart right it's time to depart so this was everything about the motion into the one day hope so you like that teaser right it was a part of the headquarter of the pw universe right now in headquarters from which we were understanding the concept say teaser is for you all it is just a different approach to analyze the physics through the visualization sometimes uh without visualization physics always used to get bored up okay so uh some students might be thinking the teaser is not having any kind of the relevancy with the content right but it is just for it is an initiative to understand the actual concept of physics what actual the physics is all about and what truly the each concept states into the form of the visualization okay guys so hope so you like the teaser i will be waiting for all of your comments beneath the video this is indirect singh signing off from the desk right again we are going to come back into the next session bye bye subscribe