hello everyone once again welcome to my YouTube channel physics with Dr sjay bav in our part one of the video Motion in one dimension We have dealt with distance displacement speed velocity instantaneous and average speed instantaneous and average velocities and certain type of important graphs relating distance time displacement time time speed time and velocity time in this video in continuation of our earlier video we'll discuss about acceleration average and instantaneous acceleration will then calculate the equations of motion which include velocity time relation displacement time relation velocity displacement relation and distance traveled in the N at second we'll also address the problem of motion under Gravity we'll also talk about relative velocity and try to discuss certain important aspects of problem solving in two dimensional problems we begin with acceleration we begin with acceleration how do we Define acceleration we Define acceleration as rate of change of velocity with time so we have uniform acceleration so what is uniform acceleration an object is set to have uniform acceleration if its velocity changes by equal amounts in equal intervals of time so in case of uniform acceleration a body must have equal change of velocity in equal intervals of time acceleration is supposed to be a vector quantity as it both has magnitude and Direction the SI unit is m/s square and the dimensions are L1 T power - 2 the acceleration can be positive negative and zero positive acceleration means velocity is increasing with time while negative acceleration also called as retardation means velocity is decreasing with time the negative acceleration is also referred as retardation or deceleration a body having zero acceleration is one where velocity is constant that is a state of uniform motion so when the velocity does not change with respect to time it's the state of uniform motion and in that case of uniform motion we have zero acceleration now what is our imagination or perspective on acceleration in case a particle has zero velocity at any instant it does not necessarily mean that it has zero acceleration at that instant what I'm trying to say is a particle can have zero velocity but it does not necessarily mean that it should have zero acceleration at that instant of time let us pick up an example when a particle is thrown vertically upward then at the highest point it has zero velocity but acceleration at that instant is equal to the acceleration due to gravity G so zero velocity is not a guarantee of a zero acceleration though velocity and acceleration are related to each other they have their own distinct meaning velocities it tells us how fast and in what direction the object is moving while acceleration tells us how its speed and direction of motion are changing so what is the information provided by the velocity it tells me how fast an object is moving and in which direction is the object moving while what does the acceleration gives it tells how its speed and the direction are changing with respect to time so both the quantities though may seem to be related but have their own individual existence if the acceleration is in the direction of a velocity that is the sign of acceleration is same as the sign of velocity then the speed of the particle increases if the acceleration is in a direction opposite to that of velocity then s of that is s of acceleration is opposite to that of velocity then the speed of the particle decreased so it tells us when does the speed of a particle increase and under what condition the speed of the particle decreases if the acceleration and the velocity have the same direction then the speed of the particle increases if the acceleration and the velocity have opposite direction then I can say speed of the particle would decrease thus sign of acceleration of a particle does not tell us whether the speed of the particle is increasing or decreasing so a very very important point to be taken care of that sign of the acceleration like acceleration due to gravity is acting in the downward direction we put a negative sign with it but that sign does not tell us whether the speed of the particle is increasing or decreasing we just just provide an example to it acceleration due to gravity XS in the downward Direction and is taken negative for a particle falling under Gravity G is negative but it increases the speed of the particle when the particle is thrown up G causes a decrease in the speed of particle so calling negative acceleration at deceleration is not just Justified just we will identify it on the basis whether the speed is increasing we call it as acceleration if the speed is decreasing we will call it as deceleration acceleration and deceleration not to be decided on the basis of the sign associated with the acceleration right a very important thing that need to be taken care of where we make certain common mistakes now we have a graph here it's a position time graph on this axis we are plotting position and on this axis we are plotting time then the graph curves upward when the direction of acceleration is same as that of velocity like in this case the direction of acceleration and the direction of velocity is the same we have curve a then the curve downwards when the direction of acceleration is opposite to that of velocity this one here acceleration and velocity are in the opposite direction so what is happening velocity is decreasing with time here velocity is increasing with time for zero acceleration the graph is a straight line which is shown here as a curve B so you have to obtain different informations under different conditions average and instantaneous acceleration let us first Define average acceleration it is defined as the ratio of change in velocity to the time interval in which the change takes place so average acceleration change in velocity upon the time interval for the change it is equal to V2 - V1 upon T2 - T1 average acceleration would be Delta V by delta T as usual this is the same way of description for average velocity and average speed instantaneous acceleration it is defined as the instantaneous rate of change of velocity with time and is the limit of average acceleration as the time interval delta T approaches 0 so instantaneous acceleration is limit delta T tending to Z Delta V by delta T can be written as DV by DT instantaneous acceleration is the time derivative of in instantaneous velocity acceleration is equal to DV by DT and V can be written as rate of change of displacement with respect to time so acceleration is also equal to d2x by DT s acceleration of a particle at any instant is the second derivative of position x with respect to time so we need to determine average and instantaneous accelerations graphically also let us calculate average and instantaneous acceleration from the velocity time graph the average acceleration during any time interval is equal to slope of a straight line joining these two points on a velocity time graph so this represents velocity time graph on the x-axis you are plotting time on the Y AIS we are plotting velocity at the point a the velocity is V1 at the time T1 at the point B velocity is vs2 at the time T2 the slope of the straight line AB is equal to Delta V by delta T So average acceleration is Delta V by delta T average acceleration is nothing but the slope of a straight line joining two points on velocity time graph instantaneous acceleration the concept Remains the Same the instantaneous acceleration at any point is equal to the slope of the tangent to the velocity time graph at that particular Point again the logic is same when the point B approaches a then delta T tends to zero in this condition the straight line AB would become tangent to the graph at Point a so limit delta T tending to 0 Delta V by delta T would be equal to limit delta T tending to zero the slope of a straight line AB which would be nothing but slope of tangent at Point a limit delta T tending to Z Delta V by delta T can be mathematically written as DV by DT which is nothing but instantaneous acceleration at Point a so what is a it is equal to slope of the tangent at Point a so we have proved here once again that instantaneous acceleration would be nothing but slope of the tangent at Point a now let us derive the kinematics equations of motion in a straight line under constant acceleration by calculus method first we pick up is velocity time relation we consider a particle moving with uniform acceleration a A can be written as DV by DT or DV can be written as as a DT at the time T is equal to 0 velocity would be equal to initial velocity U and at the time T the velocity is V Now integrate with proper limits what do we get integration DV from U to V would be integration a DT from 0 to T or integration DV is V and the limits of integration from U to V A integration of DT is once again T from 0 to T apply limits of integration V - U is = a into t- 0 so V would be equal to U + a this is known as velocity time relation what is the utility of velocity time relation it helps to calculate velocity of an object moving with uniform acceleration After Time T now the second equation of motion that is displacement time relation we consider a particle moving in a straight line with uniform velocity V so V can be written mathematically as DX by DT or DX can be written as V DT just now we have put as V isal to U + a so replace V by U + a DX can be written as U DT + a DT at T is equal 0 xal to 0 at the time T is equal to 0 X is equal to S so integrate both the sides with proper limits integration of DX 0 say s integration U DT from the time 0 to T integration a t DT from 0 to T U is constant so can be taken out sign of integration similarly a is constant so take out of the sign of integration integration of DX is X limits of integration 0 to s u is a constant integration DT would be T the limit 0 to T integration of T would be t² by 2 and 0 to T just substitute the limit it is upper limit minus lower limit so we get the equation s is equal to UT + half a² what information does it provide it provides me the displacement time relation and where it is used it is used to calculate total displacement of an object moving with uniform acceleration After Time T now the third equation is the velocity displacement equation we know a would be equal to DV by DT you can multiply and divide by DX so DV by DX is into DX by DT DX by DT is nothing but velocity so just cross multiply what do we obtain a * DX is V into DV now select the limits properly at xal to x0 the velocity is V and when X is equal to x velocity is so V it is U integrating with proper limits a DX from x0 to X U to V so a is a constant take out of the integration sign integrate DX from x0 to X similarly integrate V from initial velocity and final velocity integration of DX is X limits ranging from x0 to x v Square by 2 is the integration and U to V is the limits so it is tce a into x - x0 is v² - u² rearrange the terms and put x - x0 as s what do I get S v² would be equal to U ² + 2 a s this is is basically velocity displacement relation and is used to calculate velocity of an object moving with uniform acceleration after traveling a distance s now another important relation is distance covered in N second like we are faced with a question what is the total distance travel in the fifth second then we need an answer to it for that we'll derive a general relation to handle such situations we consider a particle moving in a straight line with uniform acceleration you would be initial velocity is is the uniform acceleration and what is SN distance traveled or covered in N seconds and S N minus1 distance traveled in nus 1 seconds let us utilize the relation s is equal to UT + half a² so in N seconds it is unu n + half a n² in N - 1 seconds it is u n -1 + a by 2 into n -1 s what is the distance traveled in the nth second it would be total distance travel in N seconds minus total distance traveled in nus 1 second just subtract the two equations rearrange the terms and we obtain here SN would be equal to U + a by 2 into 2 N - 1 so these are four important results V is = U + a s would be equal to U +/ A t² v² would be equal to U2 + 2 a SN would be equal to U + a by 2 into 2 nus 1 which would be utilized under various situations to handle the Practical and numerical problems now let us focus on motion under Gravity now motion under Gravity is a case of one dimensional motion with constant acceleration here acceleration due to gravity G is constant constant and directed vertically downward the frictional force due to air is absent that is motion takes place in vacuum so G acting vertically downward a is equal to plus G downward Direction and a is equal to if I take origin at the top then the downward direction would be taken positive if I am a at this point then a would be equal to minus G so all the signs to be decided where do we pick the origin from that is how to handle such situations we'll Define a sign convention for us which just simplifies the things for us initial of position of a body is taken as the origin for a free falling body the vertically downward direction is taken positive for a body projected vertically upward the upward direction is taken positive so under freef fall a would be plus G and for objects projected upward a would be equal to minus G motion of body falling freely under Gravity let a body fall from rest from a height edge here U is equal to 0 S would be equal to h and a would be equal to G so from this equation we obtain V isal to GT I can calculate the time of fall as V upon G from this equation U is zero so H would be equal to half GT s so the time taken would be equal to under < TK 2 H upon G from this equation v = u ² + 2 g u is 0 so v² would be equal to 2 G so V would be equal to under < tk2 G height in the N second would be 2 n -1 into G by2 assuming initial velocity U to be zero motion of a body projected upward Let the Body be projected upward with a initial velocity U and H be the maximum height attained After Time T when it comes to rest for a moment the velocity becomes zero s is equal to h and a would be minus G in this situation U would be equal to GT H would be equal to UT -/ GT s and u² would be equal to 2 G from equation 3 and 7 what do we find the value of initial and the final velocity is the same that is under root2 G it proves the speed with which a body is projected is equal to the speed with which it returns to the point of projection as U is equal to V what we can say from equation number 1 and five the time taken to go up a height is same as time taken to fall through the same height from equation number one velocity is directly proportional to time from equation number two H is proportional to t² the heights through which a body falls from Rest In Time T2 t3t are in the ratio 1 S 2 is to 2 sare is to 3 square that is 1 is 4 is 9 Square of integers from equation 3 V is directly proportional to H from equation number four hn is directly proportional to 2 N minus 1 so it answers the question the height through which a body falls from rest in first second and third second are in the ratio 1 is to 3 is 5 that is in the order of art integers this result is known as G Law of odd numbers we can also derive the equations of motion by graphical method we consider an object moving along a straight line with initial velocity U and uniform acceleration a the velocity time graph is shown here on this graph of y- axis we have Tak velocity here time initial velocity at Point a is U and the final velocity at point B is V at T is equal to 0 V is equal to U at T isal 2 t v isal to V what is the slope of this graph it would be BD upon a BC minus DC that is vus U upon T So slope of the VT graph is acceleration so I can put vus U upon t as a rearranging I get the first equation v = u + a now I need to calculate what would be the distance that would be traveled in time T it would be equal to area under velocity time graph now just look at the velocity time graph this is AB the area under this curve is equal to veloc area of the triangle plus area of the rectangle that is what we have represented as area of the trapezium o ABCO o area of triangle ABD and area of the rectangle so area of the triangle is half base into height and length into breadth substitute the physical quantities rearrange you will get S is equal UT + half a² the distance traveled in time T would be equal to area of trapezium and what is the area of trapezium half the sum of parallel sides into distance between parallel sides so that is half U + V into T So U + V would be equal to 2 s upon T also v- U is equal to a now multiply these two equations what do we obtain 2 A S is = to v² - u s rearrange I'll get the equation v² would be equal to U ² + 2 a s thus we can obtain the equations of of motion either graphically or by the calculus methods as according to our requirement hope you are enjoying the video on onedimensional motion part two once again I request you to subscribe my channel for those who have not subscribed till date all videos that you like and enjoy please comment and like the videos now moving ahead we pick up another word and that is relative velocity we have already said rest and motion are related and not absolute a given object which is at rest with respect to a particular Observer but it may be in motion with respect to another Observer the velocity is also relative and not absolute thus to describe velocity we have to mention the observe with respect to which the velocity is being measured the relative velocity of object a with respect to another body B that is v a is the rate at which body a changes its position with respect to body B we consider two objects A and B moving with uniform velocities VA and BB respectively let x0 a and x0 b be their position coordinates at the initial time T is equal to 0 x0 a minus x0 B is the displacement between the two bodies A and B at the time T is equal to Z if XA and XB are the positions of coordinates at the time T then x a would be x0 a + v a t XB would be equal to x0 B + vbt x a - XB would be x0 a - x b plus VA minus VB into p the displacement between the two bodies increases by VA minus VB into T during the time interval T relative velocity of a with respect to B is equal to displacement Upon Time and the displacement is VA minus VB upon T upon T the time cancels so relative velocity VA minus VB V would be equal to VA minus VB similarly I can calculate what would be the relative velocity B with respect to a so VBA a would be VB minus VA from equation number two 1 and 2 V AB is equal to minus of VB a the relative velocity of a with respect to B is equal to negative relative velocity of B with respect to a but the magnitude of VAB and magnitude of VBA is equal to V where what does B represent it is the relative speed between A and B the relative speed between two bodies A and B moving in the same direction with same speed with speeds VA and VB so what is V same it is equal to difference in the speed of two bodies A and B magnitude VA minus magnitude VB or magnitude B minus magnitude of va depending on on the fact whether VA is greater than VB or VB is greater than VA the relative speed between two bodies A and B moving in opposite direction with a speed VA and VB is equal to V opposite is equal to sum of the speeds of two bodies A and B VA plus VB note these equations are valid only in one dimensional motion now let us analyze relative velocity from position time graph we know that x a - XB is equal to x0 a - x0 b + VA a minus VB into T if VA and VB are equal then what will happen x a - XB would be x0 a minus x0 B if the two bodies remain at constant separation so the position time graph are parallel to each other like here they are parall to each other since the velocities are equal so the displacement at the time T is equal to Z and at a later time is equal then we may have we may VA may not be equal to VB then one graph would be more steeper than the other they meet at a point for example let us pick up VB is greater than VA then what will happen just look here VB the slope is more and VA the slope is less they meet at this time T so from the initial time T is equal to 0 to T Das the separation decreases and after T Das the separation once again increases so we see the graphs meeting at Point P at the time T Das at T Das body B overtakes body a before T Das XA - XB is positive and after T Das XA - XB is negative now we'll handle certain typical problems that you come across in two-dimensional motion we have an example one a man has to go 50 m due north 40 m due east and 20 M du out to reach a field what distance he has to walk to reach the field what is the displacement let us draw the figure for it so it says 40 m to the east what is our concept of Direction This is north south east and west so 40 m east 30 m North and 20 M sorry 40 m to the east 20 m to the north sorry 50 m North 40 m due east and 20 m due south so we are here 20 m to the South downward Direction 50 m to the north upward Direction just arrange the vectors now actual distance traveled is from A to B B to C and C to D so it is 50 plus 40 + 20 giving me 110 M when it comes to the displacement then we have to use polygon law so it would be a vector plus BC Vector plus CD Vector 50 JC + 40 I + 20 minus of JC it is 40 i + 30 JC so magnitude of the displacement would be 50 m if I need to calculate direction of the displacement then direction of the displacement would be calculated by tangent Theta which would be given by from the figure de upon AE which is 3x4 and tangent inverse 3x 4 would be 37° so we have magnitude of the displacement as 50 m and the Direction 37° with respect to the horizontal now let us pick example number two a man walks at a speed of 6 km/ hour for 1 km and 8 km/ hour for next kilometer now average speed is total distance upon total time S1 + S2 upon T1 + T2 total distance is 2 km time taken to cover S1 distance 1X 6 hour and time taken to cover S2 distance is 1x 8r so just put in the relation S1 + S2 upon T1 + T2 so this would be 1 + 1 2 and 1X 6 + 1 by 8 which on simplification gives 48x 7 km/ hour so that's the average speed given by the harmonic mean of the two speeds if the speed of the particle is V = 10 t² m/ second then find out the distance covered between t = 2 to tal 5 Second we have V isal DX by DT so DX would be equal to V DT integrate both the sides with proper limits so DX from 0 to X and vdt from 0 to 2 to 5 so integration DX is X integration V can be replaced by 10 t² t² integration is T Cub by 3 so we have 10 by 3 5 Cube - 2 Cube and the total distance traveled is 390 m then we have the example number fourth the displacement time graph of two objects A and B is shown in the figure find the ratio of the velocities of the object velocity of the object would be slope of the displacement time graph that is tangent Theta so velocity of object a upon velocity of object B VA upon VB is equal to tangent of theta a upon tangent of theta B Tang 30 upon Tang 60 Tang 30 is 1 byun3 and T 60 is < tk3 so that is 1 by 3 so the ratio of the velocities VA and VB is 1 is 3 VA upon VB is 1 is 3 example five calculate the displacement and distance traveled from the graph so area under velocity time graph would provide us a measure of displacement and distance distance travel is area of the first Square second square and the third one and let us calculate that area the first area would be equal to 4 into 2 second area would be 2 into 2 plus third area would be 2 into 2 that is 16 M but displacement is a vector and in that case it would be area of the first minus area of the second because it is below the x axis and then the area of the third one so that gives me 4 into 2 - 2 into 2 + 2 into 2 and the total displacement is 8 m so distance traveled is 16 M but the displacement traveled is only 8 m example number six a jet airplane traveling at a speed of 500 km per hour ejects its product of combustion at a speed of 1,500 km/ hour relative to the Jet Plane what is the speed of the L with respect to an observer on the ground so what are we given velocity of jet plane with respect to the ground we call it as velocity of plane with respect to ground then velocity of the combustion products with respect to the plane so we can write what would be velocity relative vpg vpg would be VP minus VG vcp would be VC minus VP we are required to calculate velocity of the combustion product with the respect to the ground vcg that is VC minus VG the combustion product and the plane are moving in the opposite direction so equation 1 plus equation number two so we get vpg plus vcp so this VP cancels and we get VC minus VG which is is nothing but vcg the required one so vpg it is 500 vcp has opposite direction to the motion of the plane so we write it as-500 so the net velocity is 1,000 km/h so velocity of combustion products with respect to the ground is 1,000 kilm per hour example seven a policeman is moving on the highway with a speed of 30 kmph it fires a bullet at a thief's car speeding away in the same direction with a speed of 192 km/ hour if the muzzle speed of the bullet is 150 m/s with what speed does the bullet hit the thief's car so speed of the police van is 30 into 5 by8 that is 25 by 3 m/s speed of the bullet is 150 m/s what is the effective speed of the bullet that is V1 would be equal to velocity of the polican and speed of the bullet that is 475 by3 m/s the speed of the thief's car is 192 into 5 by8 that is 160 by3 m/s since bullet and the theves car are moving in the same direction so what is the relative speed between bullet and the car it would be V1 minus VC which would be equal to 105 m per second this is what is the speed with which the bullet would hit the car let us pick another example example number eight two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minute a man cycling with a speed of 20 k km/ hour in the direction A to B notices that a bus goes past him every 18 minute in the direction of his motion and every 6 Minute in opposite direction what is the speed T what is the period T of the bus service and with what speed do the buses fly on the road so let us see speed of the bus we represent by VB speed of the cyclist VC and VC is given as 20 km/ hour distance covered by the bus in t minutes is VB into T relative speed between bus B1 flying in the same direction so V same would be VB minus VC distance covered by the bus B1 relative to the cyclist in 18 minutes is V same into 18 for the bus B1 to pass the cyclist what is required 8 into V same 18 into V same is equal to VB into T and we call it as equation number one now similar situations for the bus second moving in opposite direction relative speed of bus V2 apping opposite to the motion of cyclist and the cyclist V opposite would be VB plus VC distance covered by the bus B2 relative to the cyclist in 6 minutes it would be equal to V opposite into 6 for the bus to pass the cyclist 6 into V opposite must be equal to VB into T from equation number one and two 6 into V opposite would be 18 into V same substitute what do we get VB + VC would be equal to 3 * VB minus VC so VB would be equal to twice VC so 2 into 20 that is 40 km per hour what is time period T it would be 6 V opposite upon VB which would be 9 minutes so the time period is 9 minute and the velocity of the bus is 40 kilm per hour hope our two videos of onedimensional motion part one and part two have clarified the basic concepts of onedimensional motion if you have all enjoyed and understood the video you are all requested to subscribe the channel if you have not done and you are requested to like and comment on the videos that you are observing to enhance your knowledge thank you see you in the next video