Kinematic Equations of Accelerated Motion

hi friends ready to learn the chematic equations of accelerated Motion in this video we are going to look at the concepts of instantaneous velocity instantaneous acceleration and equations where acceleration is not constant we are going to use the powerful method of calculus to look at these equations and we&#39;ll also use the calculus method to derive the equations of motion I&#39;m going to make these Concepts really easy for you and explain you with some simple examples so make sure you watch the video till the end and do check out the full courses on our website and Android app Manoa Academy I&#39;ll put the links below and the best part is you can try out these courses for free so do check it out links are given below all right let&#39;s dive into kinematic equations I&#39;m sure you&#39;ve seen these equations before what are these equations these are the equations of motion and remember the symbols here U is initial velocity V is final velocity T is time s is displacement and a is acceleration so these are equations of motion for an accelerated motion but there&#39;s an important condition here the acceleration needs to be constant so the a that you see in these equations represents acceleration and that must be constant which means that the velocity should be increasing or decreasing at a constant rate but what if we have a motion where the acceleration is not constant what do you do then that&#39;s exactly what we&#39;re going to look in this video because we will look at the kinematic equations for Accelerated motion and we&#39;ll use the mathematical concepts of calculus and look at these equations and I&#39;m going to make these Concepts really clear for you so let&#39;s dive right into it let&#39;s start with the concept of velocity what is velocity remember velocity is defined as the rate of change of displacement and the formula is displacement divided by time so if a body has a certain displacement or a change of displacement you can denote that as Delta s over a certain time interval which we can denote as delta T so the change in displacement by change in time is the formula of velocity so let&#39;s take an example let&#39;s say a body travels a distance of 10 m so the displacement is 10 m here and let&#39;s say the time interval is 5 seconds so we have me measured its displacement over 5 seconds so what will be the velocity of the body very simple change of displacement by time so 10 m by 5 Seconds it&#39;s going to be 2 m/s now we have measured this over a 5sec time interval now within that time interval maybe the body was changing its velocity right maybe it was going a little faster a little slower just like when a car travels on the road it doesn&#39;t perfectly travel at a constant speed right or at a constant velocity so maybe it covered this 10 m in 5 seconds but the velocity was changing so what have we really measured that&#39;s right we have measured the average velocity the average over this 5sec time interval so remember this important concept that this formula is really the average velocity over your time interval and the average turned out to be 2 m/s but now let&#39;s say we want to find the speed at a particular instant like in the cars you have seen there&#39;s a speedometer right so it shows the speed of the vehicle at that instant not over an average time because whenever you&#39;re looking at this uh speedometer let&#39;s say the car is moving at a very high speed right or your bike is moving at a high speed you will see it shows 60 km/ hour 80 km/ hour when you are stopped it will show zero when you&#39;re going slow 10 km/ hour so you have seen the speedometer constantly changes and it is giving you the magnitude of this velocity the speed at that instant so that is what we call instantaneous velocity and how will we get that to get the instantaneous velocity we must make this time interval very small we cannot meas measure it over a long time interval otherwise that will become an average you need to measure it over very short interval like1 second or 01 second or 01 second make that time interval very small so in physics or mathematics we write it as limit delta T tending to zero so when this time interval tends to zero that is denoted as DS by DT so here we are using the mathematical tools of calculus here where the short time interval we denote it as DT so DS by DT is basically your instantaneous velocity where s is displacement and T is time so you&#39;re doing a differentiation with respect to time of the displacement so remember this thing that your instantaneous velocity your instantaneous velocity is going to be DS by DT so let&#39;s write down that important formula V = to DS by DT and you have done graph stuff also right so we can say that if you draw draw a graph of displacement versus time velocity is basically the slope of that graph because when you do DS by DT or delt s by delta T you&#39;re finding out the slope remember slope was y by X or change in y by change in X so basically velocity is the slope of the displacement time graph and you can calculate it as tan Theta at a particular point but we&#39;ll talk about slope Concepts in a separate video in this video let&#39;s focus on these equations and calculus stuff to understand the concept and formula let&#39;s look at an example so here we&#39;ve been given that displacement of a moving body is given by this equation S = 5 T ^2 + 4T so we can know the relation between displacement and time and what do we need to find here we need to find the instantaneous velocity at time T = 2 second and also the average velocity so let&#39;s first start off with the instantaneous velocity so guys what is the instantaneous velocity formula remember V = to DS by DT so we are going to use this formula V = DS by DT for instantaneous velocity always use DS by DT Delta s by delta T is for average velocity right so here I can get the instantaneous by substituting s over here so we will do DDT that is a differentiation of the displacement which is 5T sare + 40 and you guys know how to do this this is all calculus stuff so please revise your Calculus because it is very very important and very useful for this chapter so we have to do a difference of this displacement and we know that we can do differentiation of each of the terms separately so this is basically going to be become DDT of 5 t² right plus DDT I&#39;ll break it down of 4T now you guys know how to do the differentiation you bring the constant outside so this is finally going to become 5 DDT of t² plus 4 * DDT of T now what is DDT of t² in calculus you know this formula that DD X of x ^ n is n * x ^ n -1 so here my variable is not X it&#39;s t so let&#39;s apply that so DDT of t² will be become n * T ^ N - 1 here n is 2 so 2 * T ^ 2 - 1 which is 1 so 2 * T so use that formula and you will basically get 5 * 2T here plus 4 * what is DDT of T that&#39;s simple it&#39;s one again you can apply that formula it will be 1 * T ^ 1 - 1 which is T ^ 0 so again 1 so this is going to be + 4 so here we have got our instantaneous velocity V is 5 * 2 so that is 10t + 4 so that is my expression for instantaneous velocity so that is the general expression in terms of T but the question said find the velocity at time t equal to 2 second so it&#39;s very easy all you have to do is substitute tal to 2 second here done so what is the velocity at 2 seconds we can say at T = 2 second the velocity is going to be V = 10 * 2 + 4 so what is the answer 20 + 4 24 and guys don&#39;t forget the units 24 m/s actually here we should have been given the units and meters you know so that is missing in the question here that the displacement should have been given in meters since time is given in seconds uh but if the units are not given then you&#39;ll have to write 24 units okay whatever the unit is kilometer per hour meter per second so there you can see that is our final answer here so that&#39;s all you have to do if they ask you instantaneous velocity do the differentiation do DS by DT and if they ask you at a particular time just substitute it and you&#39;re done next let&#39;s look at the second part which is average velocity between 2 seconds and 3 seconds now let&#39;s calculate the average velocity it&#39;s been asked to find the average velocity between time 2 seconds and 3 seconds so for that we&#39;re going to use the average velocity formula so my average velocity formula is Delta s by delta T change in displacement by change in time so we can write that as S2 - S1 the change in displacement divide by change in time T2 minus T1 now what is the time here T1 is 2 seconds and T2 is 3 seconds right that is the time which has been given here and what is the displacement at these times the displacement we can get using this formula S = 5 t^2 + 4T you just have to substitute the time over there very simple so let&#39;s work it out so S2 which means the displacement at 3 seconds is going to be 5 * 3 2 + 4 * 3 minus the displacement at 2 seconds so just substitute T = 2 there so 5 * 2 2 + 4 * 2 so that&#39;s all we&#39;ve just substituted the time values in the displacement expression div divided by the t2 - T1 3 - 2 so simple there you go so now you just have to do your calculations here so let&#39;s go ahead and do that what will this be 5 * 3 2 that means 5 * 9 this is going to be 45 + 4 * 3 is 12 minus uh this we need to be careful about this entire thing should be in minus right in uh should be under the brackets over here because we are doing S2 - S1 so 5 * 2 2 is 5 * 4 so that&#39;s going to be - 20 and we have minus 4 * 2 is 8 over there divided 3 - 2 is 1 so just calculate what will you get over here you&#39;re getting 45 + 12 so which is basically 57 - 28 / 1 so the answer is 29 m/s this is our average velocity average velocity 29 m per second and that is from the time 2 to 3 seconds so see it&#39;s very simple just use the formula and you can get the average velocity and just carefully substitute the values calculate and there you get your answer so we have seen how to find velocity from displacement now how do you find displacement from velocity so we know velocity is displacement by time so you can rearrange the equation and you&#39;ll get displacement equals velocity multiplied by time so simple but this equation is representing your average velocity so if you know your average velocity you multiply by the time you will get the total displacement in that time interval right so total displacement is average velocity multiplied by time but what if the velocity of the body is changing over the time and you don&#39;t know the average velocity so let&#39;s say we want to work with the instantaneous velocity then what do we do so remember for instantaneous velocity how do we calculate it from the displacement the formula was the instantaneous velocity so I&#39;ll write I N St that is for instantaneous velocity equals DS by DT so to calculate it what do we do we do a differentiation with time we do DS by DT so you can see here we are doing a difference here we are doing a differentiation this DS by DT is the differentiation in calculus and that way you can find out the instantaneous velocity the velocity at a particular instant now suppose we know the velocity at certain instances right we know the instantaneous velocity and I want to find the displacement from that so how do I do that so let&#39;s see if we can use this formula so if you look at this formula I&#39;m interested in calculating displacement so I can rearrange this and write it as if I bring the DT here so we can do V * DT = to DS since we interested in displacement let&#39;s bring that to the left hand side so or I can write DS = to vdt what is the meaning of this equation that in a very small time interval if you multiply the instantaneous velocity you will get that small instantaneous displacement in that small time interval but we want to find the displacement over a certain time so what do you have to do we have to add up all of these so vdt so the vdt values need to be added up across that time interval and how do you do that so here again we use the calculus method of integration because finding that area right adding up all these vdts finding the summation here of all this is what we represent in calculus as integration because we want to find the total displacement so if we integrate both sides if we do integration of DS equal to integration of vdt what will we get therefore we get S because when you integrate displacement you get displacement it will be the integration of vdt so you can see so you can see it is exactly the reverse process to get instantaneous velocity from displacement we do differentiation V equal to DS to get displacement from velocity naturally you&#39;ll do the opposite thing so instead of differentiation you are going to do integration here and integration of vdt over the time interval will give you displacement now you might be thinking what is the time interval that will be given to you so normally you&#39;ll see in the books they say over a certain time interval 0 to T or you can write T1 to T2 so that will end up giving you the displacement so remember this very important concept and here again we are talking about instantaneous velocity right and the total displacement calculated from that that is why we are using all this calculus stuff dsdt and this integration of vdt don&#39;t get scared by this calculus terminology I know initially it sounds confusing and scary but it is just a transformation of the basic concept that you have learned we just applying calculus here and we are getting these formulas clear so let&#39;s understand how we can calculate displacement from instantaneous velocity using this example so here we have been given that velocity of a moving body is given by V = 10t we need to find the displacement from 0 to 4 seconds so here you can see the velocity of the body is constantly changing because as the time changes it is 10 * T okay so at time equal to 0 it is 10 * 0 0 at time let&#39;s say at 1 second it becomes 10 m/s at 2 second it becomes 20 m/s at 2.5 it will become 25 m/s and so on so clearly you can see velocity of the body is constantly changing and this is the instantaneous velocity this gives us the formula of the Velocity at any time T so how do we find the displacement from 0 to 4 seconds so here once again we have to use that integration concept because we have been given instantaneous velocity and what was our formula to find the displacement from instantaneous velocity you need to integrate the velocity over the time interval so we have to integrate V DT okay and the time interval here has been given from 0 to 4 seconds so put the limits from time 0 to 4 seconds this is the integration and that will give me the total displacement that&#39;s it so go ahead and substitute this 10t into the formula so what will we get from integration from 0 to 4 seconds and this is 10t the velocity is 10t DT now how do you do this integration whenever there&#39;s a constant you bring it out so this will be integration from 0 to 4 seconds the 10 comes out T DT now what is the integration formula and calculus when we do integration it is integration of x ^ n DX is going to be x ^ n + 1 / n + 1 this is the formula integration is the opposite of differentiation there you had n minus 1 here there is n + 1 right that was n * x ^ nus1 so here what is our power over here you can see this is T that means this is basically T to to the^ 1 so we just need to substitute n = 1 over here so basically what happens so this will give us 10 times and integration of T DT will be T ^ 1 + 1 t² / by 2 so we are getting t² by 2 and the limits here are from 4 to 0 so the limits for this are from 4 to 0 so what does that mean that I need to substitute four here and 0o and then subtract them right because when you have limits you basically subtract here so it&#39;s going to be 10 * T ^2 by 2 so with the value 4 here so that&#39;s going to be 4 2 by 2 and then you substitute Min - 0 2 by 2 right and so what do we end up getting this will become 10 and this 4 S by 2 is basically 16 by2 that&#39;s 8 8 - 0 and this is going to be 8 * 10 which is 80 M 80 m is your answer again uh this should have been given all in SI units so it should be mentioned here you know these are all in SI units and your answer is going to be 80 M for the displacement so remember this important way of doing integration and finding the displacement and also you guys should know that when you are doing this integration when the limits are not given it is called an indefinite integral then we put a constant over there plus C but when the limits are given then this constant is not there so that is why I did not add the constant here because the limits were given to us from 0 to 4 seconds if that was not given then you need to add a plus constant that is indefinite integration indefinite integral and this was a definite integral because the limits were given to you we substituted that and there you can see we have got our answer now let&#39;s accelerate to acceleration what is acceleration the change in velocity divided by time because it is defined as the rate of change of velocity so the change in velocity by the change in time we can also represent that in a formula like this acceleration a is Delta V that is the change in velocity divided by the time interval delta T and you&#39;ve all seen this formula right Delta V if you represent as final velocity minus initial velocity so vus U / T so let&#39;s say the time is from 0 to T but now what if the acceleration is changing right what is is the acceleration at a particular instant just like we had instantaneous velocity how can we Define instantaneous acceleration because once again this will give me the average acceleration over a large time interval so what this is really is really your average acceleration because acceleration need not be perfectly constant all the time that will become a case of uniform acceleration so how do I Define that or what is the formula going to be so again if you want to measure the the instantaneous acceleration in a very very very small interval of time what do we need to do again we&#39;ll use calculus we&#39;ll say in the limit delta T tending to zero that means a very small time interval my instantaneous acceleration instantaneous this is a very interesting word instantaneous it has all the vowels see a e i o U right in fact I think I just noticed that while writing I hope I got my spelling right it&#39;s a long word so instantaneous acceleration is going to be a will be when the limit tends to zero we denote it as DV by DT this means very small time interval so acceleration is DV by DT this is the formula for instantaneous acceleration and once again you&#39;ll have that slope concept right so if you&#39;re having a velocity time graph acceleration is DV by DT so it is basically the slope of the Velocity time graph so remember these formulas this is the formula for average acceleration and instantaneous acceleration you&#39;re doing a differentiation of the Velocity with respect to time best way to understand formulas is with the help of an example so let&#39;s take a look at this example here we have been given displacement of a moving body s = 5 t^2 + 4T and we need to find the instantaneous acceleration of this body so how will you do it remember the instantaneous acceleration formula instantaneous acceleration a is DV by DT we do a differentiation of the Velocity but what has been given here is displacement we have been given S no problem we are going to first find velocity and then plug it in here so we can start off with this relation V equal to the instantaneous velocity V equal to DS by DT so let&#39;s plug in this uh displacement over here so we have DDT and the displacement is 5 t² + 4T so go ahead and do a differentiation of that so if you do DDT of 5 t² it&#39;s going to be 5 DDT of t² + 4 * DDT of T so what will you end up getting DDT of t^2 like we saw is 2 * T you can use that n * x ^ nus1 formula so this will give you 5 * 2T and DDT of T is 1 you can easily remember that like this right DD DT and DT seems to be cancelling so that&#39;s going to be + 4 so I&#39;m getting the 10t + 4 in fact we had seen this earlier when we had calculated the instantaneous velocity now we need to plug this in here so what will we get the instantaneous acceleration will be DDT and we are basically plugging in the Velocity in this formula so our velocity is 10 t + 4 the instantaneous velocity so do a differentiation here what will you get again 10 * DDT of t plus DDT of 4 four is a constant so its differentiation will be zero so therefore finally we are getting DDT of T is 1 so 10 is the answer acceleration is 10 if units are not given you can simply write 10 units if these are given in seconds and all you know or SI units you know it&#39;s going to be m/s squar so here in this case the instantaneous acceleration value turned out to be constant so it is actually a case of uniform acceleration but it need not be maybe you could have got an expression like 10t or you know 5T or something like that so this is what we have got here so basically you can see you can find acceleration as differentiation of the Velocity if displacement is given it&#39;s like a double differentiation so if you think about it a is DV DT and if we substitute v as dsdt so this is like a double differentiation so basically what you&#39;re doing a DDT squ of your displacement so this is another way to look at acceleration or you can start off with this and do the steps and get it and remember here they just said instantaneous acceleration they didn&#39;t ask you to find acceleration at a particular time so when they give a question like this you can leave it in terms of T suppose we got 10t over here you leave it as that because they have not given you that I want to know the acceleration at this time at 4 seconds 5 Seconds nothing like that is given so you can leave the expression in terms of T like that we have seen how to calculate acceleration from velocity but how do you calculate velocity from acceleration so of course you can rearrange the simple formula and you will get velocity is acceleration multiplied by time time but the Restriction here is that acceleration should be constant so for uniform acceleration this works great you can get the velocity but what if the acceleration is changing like in the case of instantaneous acceleration so once again we have to use the concept of calculus remember how did we Define instantaneous acceleration instantaneous acceleration is the differentiation of velocity so DV DT so to find acceleration from velocity what were you doing differentiation with respect to time so we were using differentiation so I think you must have guessed that how do you calculate velocity from acceleration you have to do the opposite of differentiation which is basically integration and that you can easily see from this equation because if you rearrange it so if you multiply the take the time this way so you get a DT equal to DV since I want to find velocity I&#39;ll bring it on the left hand side so I can say DV equal to a * DT what does this mean that if you multiply acceleration with a very very small time interval you will get the value of that small velocity and to get the full velocity you need to take a summation here right summation in statistics we represent with Sigma but in calculus we write that big kind of of that kind of s symbol for integration so what you really need to do is if you want to calculate velocity from acceleration you need to integrate both sides we can integrate both sides of the equation so on integrating both sides integral of DV will give you the velocity that&#39;s our goal here and this will be the integration of acceleration time DT again remember if uh the time interval is not given then obviously you it&#39;s a indefinite integral you have to take plus a constant but if the time interval is uh given then you directly substitute the values here so let&#39;s say you want to calculate the velocity from for the time interval 0 to T so your integration formula is going to look like this so remember this important concept that to get instantaneous acceleration from instantaneous velocity you do DV by DT but if you want to get the velocity the instantaneous velocity then you have to do integration of the acceleration so you have to do exactly the opposite and to get velocity from acceleration this time you need to do integration so very very important Concepts please understand it and write it down so let&#39;s see how we can calculate velocity from acceleration using this example so what have we been given here acceleration of a moving body is given by a = 3T ² and we need to find the velocity at Time 4 seconds if initially the body is at rest so some condition has also been given to us so once again let&#39;s go ahead and apply our formulas of calculus and see what we can get here so our goal is to find velocity from acceleration the instantaneous acceleration is given to us a equal to 3 t² acceleration is dvdt but velocity is going to be the integration the opposite right so if I want to calculate velocity it is going to be the integration of the a DT right and if a time interval is not given then it&#39;s an indefinite integral like this so let&#39;s go ahead and substitute the acceleration expression here it&#39;s going to be integration of 3T Square DT so what will that be guys you bring the three outside it&#39;s a constant integration of t² DT and once again here&#39;s use your maths formula that integration of x ^ n DX is going to be x ^ n + 1 by n + 1 and if it&#39;s indefinite you will have a constant over there plus C so since no time interval is given I&#39;ll I have to take that constant plus C here so what will be integration of t² T Cub by 3 right so this is going to give me 3 * T Cub by 3 plus a constant C and this three and three will cancel here so what are we left with TQ + C so velocity is going to be equal to TQ + C this is my expression and I need to find the velocity at tal to 4 second so if I simply substitute four over here I&#39;m still left with this plus C this constant is irritating over here right I don&#39;t know its value but we can find out its value because we have been given a condition that initially the body is at rest initially means time equal to zero so for this special condition given to us at time equal to0 we know that velocity is zero so see using this can you calculate the constant of course you can so if you substitute it here what will you get put V = to 0 t = to 0 so 0 CU plus C so therefore you&#39;re getting your constant C is Zer so that&#39;s it we know the constant is zero so finally we can get velocity at 4 second so at T = 4 second V = TQ Q which means 4 CU which means 64 again units right or if the it&#39;s in SI units then it will be 64 m/s so here we are all assuming SI units here so your final answer is going to be 64 m/ second 4 Cub so see just apply your integration and again remember if limits are not given then you need to take a plus constant if limits were given to you then you&#39;ll have to just simply subtract suppose the limits were given let&#39;s say 2 to 3 then you just have to do 2 uh TQ uh from the value at 3 minus the TQ value at 2 so it will become 3 Cub minus t CU But Here No Limits were given so we took a plus c and easily calculated our answer here here are the three famous equations of motion and as we discussed what is the restriction here the acceleration a is constant so these are all uniform acceleration where the acceleration is a constant value now you must have studied how to derive these equations of motion using algebraic methods and using graphical methods but here we&#39;re going to do something very interesting we are going to derive it using calculus so we will start with those instantaneous velocities or uh instantaneous acceleration all those Concepts that we learned with all this differentiation integration and we are going to apply all those equations to derive these equations because remember the equations that we&#39;ve been looking at using calculus is for any general motion right the velocity does not have to be constant the acceleration does not have been to be constant because it deals with instantaneous velocity instantaneous acceleration so those equations are valid in the general case and let&#39;s see if we can apply that to get our special equations of motion where acceleration is constant so we&#39;ll start with the derivation of the first equation of motion here let&#39;s look at the derivation of the first equation of motion v = u + a this time using calculus technique so we will start with the equation of acceleration the calculus equation for instantaneous acceleration we know a = DV by DT and the main trick about these derivations is knowing the starting step where to start after that you&#39;ll see the flow is easy so we know acceleration is dvdt now in this equation you can see the goal is to write velocity the final velocity V is initial velocity U Plus 8 right we want to uh derive this equation so clearly we want to find V in terms of a so what do we do we&#39;ll rearrange this equation as we&#39;ve done before and so we&#39;ll write DV equal to a DT and we know that if we integrate both sides we should get the velocity so I can apply integration on both sides right now what will be the limits of integration here in this equation you know we take the body is going from a time interval 0 to T so we can see that the integration for the time will be done in the limits 0 to T and the velocity at Time Zero we denote with u that is the initial velocity so here the limit will be U and at time T we call it V so you can see those are the limits for our velocity so now go ahead and do the integration and see if you&#39;re able to get that equation so when we do the integration on the left hand side we have DV so integration of DV is going to be V and then you apply the limits here from U to V initial to final velocity what do we have on the right hand side on the right hand side we have integration a DT but remember in this the acceleration is constant that is the Restriction of the equations of motion it is a constant acceleration the rate of change of velocity is constant velocity is not constant the rate of change of velocity which is the acceleration is constant since it is a constant term I can bring it outside right of the integration so what will I get a comes out and we have integration 0 to T of DT so on the right hand side we&#39;ll have basically integration of T will give you integration of DT will give you T and we apply the limits here from T to zero and here you have V with the limits V to U now what is the meaning of this limits basically you set this variable equal to V and U and take the difference so it is basically left hand side becomes V minus U you substitute V and minus the lower limit upper limit minus the lower limit same way over here upper limit minus the lower limit you substitute into T so it&#39;s going to be a * T minus 0 now go ahead and simplify this this is very easy v- u = a t and if you rearrange it what do you get v = u + a and there you go we have our first equation of motion and yes we have derived it using calculus so we just applied simple integration here starting with the uh formula of acceleration or you can start off with DV equal to a DT right and there we have got our first equation of motion now let&#39;s take a look at the second equation of motion now let&#39;s look at the derivation of the second equation of motion which is s = UT + a² remember s is displacement U is initial velocity T is time and a is acceleration so we can see we are clearly interested here in displacement there&#39;s velocity time acceleration so let&#39;s start off with the velocity formula V = dsdt since we want to calculate displacement let&#39;s bring that on the left hand side so we&#39;ll bring velocity and time on the other side because you can see in the equation of motion displacement is on the left hand side here so I can write DS equal to vdt right DS equal to V DT now here we could do an integration but you can see that the second equation of motion doesn&#39;t have V the final velocity it has U the initial velocity so this is where you can apply your first equation of motion V equal to U + a and change this in terms of U and a so what you&#39;re going to do is substitute V = to U + 8 from the first equation of motion we are allowed to use the first equation because we have proved it okay so we are getting DS equal to U + a * DT now I want to find the displacement that means we need to do an integration of this expression because I&#39;ve got DS the small displacement we need to add it up so we will do an integration on both sides and this will get me the displacement from 0 to S and here the limits will be 0 to T So once again go and apply your Calculus fundas and do this integration so what will you get here if you integrate DS on the left hand side we&#39;ll get S with the limits s and Zer okay and what do we get on the right hand side when we are integrating this so we will get this expression is U + a * DT so when we are integrating it we will get integration of from 0 to T of U DT plus integration of 0 to T of a a t * DT you can break the plus down like this so let&#39;s go ahead and do that integration here initial velocity is a constant number so we&#39;ll bring it out so this becomes U integration from 0 to T of DT acceleration is constant in the equations of motion so we can bring that out acceleration comes out here integration from 0 to T of T DT so what will you get over here integration of DT is T and you have the limits T to zero and here we have a integration of T DT is going to be T ^2 / 2 so what we have over here T ^2 by 2 again the limits of T and z and on the left hand side we have S with the limits s and Z so go ahead and substitute the limits and you&#39;re done so the left hand side basically becomes s- 0 which is s or you can write s - 0 here this becomes u u t minus 0 is t plus half a t² since again it is from T to 0o so we&#39;ll put t² here minus 0 will of course give you Zer so there you see we are getting s = u + half a [Music] t² and we are done we have derived the second equation of motion using our calculus method again the main trick is to know how to start and you go ahead and do the calculus method you apply your integration and you will easily get the derivation here now let&#39;s look at the derivation of the third equation of motion by calculus method the third equation of motion is v² = U ^2 + 2 a s we have acceleration a in here so using calculus method we can first start with the formula of acceleration what is the formula of instantaneous ACC acceleration instantaneous acceleration a is DV by DT but if you just try to integrate this equation there&#39;ll be a problem because we don&#39;t want the time variable there&#39;s T in here but the third equation of motion does not have any time so we have to do some magic and get rid of this time and what we really need is a v they are fine and we need displacement s so what can we do here let&#39;s take a look so what you can do you can change this into to get rid of time you do DV by DS and since we are dividing by DS I&#39;ll again multiply by DS so that it cancels and I&#39;ll shift the DT here so write down this is the magic step to get rid of the time variable how do we get rid of the time because what is DS by DT s is displacement so DS by DT is rate of change of displacement that means it is velocity so this part so this term here is simply velocity V so what can we write this as we can write acceleration a is DV by DS multiplied by the velocity V here so see time has been kicked out time is not there now what do we do just cross multiply bring the DS this side and let&#39;s do some maths integration so what will we get here so we can can do a DS equal to V DV okay and now to get that equation let&#39;s integrate on both sides so when you integrate on both sides so what will be the integration limits here so let&#39;s say initially the displacement is zero at the initial point right and the final displacement is s and when I integrate velocity the initial velocity we denote by U and Final velocity by V so these are my integration limits now go ahead and do the integration for these equations of motion you know the important thing that acceleration is constant so since a is constant we can bring it outside the integration so what will this become a and integration of Simply DS here we have integration of V DV so inte ation of V DV is going to be v² by 2 using that integration formula and the limits will be from U to V and when you do integration of this what will you get you will simply get S so integral of DS will be S and the limits will be from 0 to S so now just plug in the limits and what do we get a * s - 0 equals we&#39;ll plug in these limits here v² - U ² or v² by 2 - u s by 2 so you can have two in your LCM now just cross multiply and see what you get so if you bring the two over here we are going to get 2 a s on the left hand side and right hand side will be v² - U ² you can see we have got to the third equation of motion just rearrange it and and you will get v² = U2 + 2 a s there you go we have derived the third equation of motion using calculus method the main trick again is how to start we started with the acceleration formula and this was the magic trick you have learned in fact you can see there&#39;s a sort of a new formula you have here a = v DV by DS or a = d DV by DS * V and we use this to integrate and got our third equation of motion so simple here&#39;s our formula summary that will help you revise what we have learned so remember we started with these three equations of motion 1 2 and 3 and you have done these before in class N9 right V = to U + a s = to UT +/ a s and v² = to U2 + 2 a s these are the three famous equations of motion but they have a condition what is the condition even though they are accelerated motion the acceleration should be constant so here the a the acceleration in these equations of motion is constant it is a fixed value for example 5 m/s Square 10 m/s square whatever be the value it is fixed so it&#39;s accelerated motion where velocity is changing but acceleration is fixed then you can use these equations but if acceleration is changing right or we have this instantaneous velocity instantaneous acceleration then what do we do then please don&#39;t use these equations you need to use these equations these are the equations for this instantaneous velocity instantaneous acceleration so I&#39;m going to write that word over here instantaneous so that you can instantly remember what to use and what is the instantaneous formula V equal to DS by DT remember average velocity was Delta V by delta T So instantaneous velocity is DS by DT similarly acceleration was Delta V by delta T instantaneous acceleration is DV by DT so what are we doing we are doing a differentiation here with respect to time we using calculus so you can get velocity from displacement by doing differentiation how can we get the displacement from the velocity how do we do that you just do the reverse you do integration reverse of differentiation is integration so the formula is s equal to integration of v d v DT it&#39;s easy to remember just take the DT this side and integrate same way just like you can get acceleration from the velocity by doing differentiation you can get velocity from acceleration by doing integration just the reverse so I hope this formula summary will help you remember these important Concepts and do write down these formulas the best way to remember formulas is to write them down not just look at them okay so go ahead and jot them down and I&#39;m sure you&#39;ll remember them now I have an interesting question for you the question says displacement of a moving body along the x- axis is given by x = 60 + T Cub so the displacement is given in terms of time you need to find the acceleration at time equal to 5 seconds so go ahead and try this question and let me know your Answers by putting it in the comments below and I promise to reply to your answers so go ahead apply the concepts that you&#39;ve learned and try out this question and let me know your answers by putting it in the comments below friends I hope these kinematic equations of accelerated motion are crystal clear to you now and remember practice makes you perfect so go ahead and apply these Concepts and practice more questions on this topic and if you like this video hit the like like button and share it out with your friends and if you haven&#39;t subscribed to our YouTube channel come on hit that subscribe button and click on the notification Bell so that you don&#39;t miss out on any of our videos and do check out the full courses on our website and Android app I&#39;ll put the links below and the best part is you can try these courses out for free we have courses on physics chemistry biology maths coding and artificial intelligence so do check out these courses because in these courses you&#39;re going to get live classes interactive videos quizzes questions mock tests and revision notes so do check out our website Manoa academy.com links are given below and share it out with your friends and stay connected with us and just keep learning

hi friends ready to learn the chematic equations of accelerated Motion in this video we are going to look at the concepts of instantaneous velocity instantaneous acceleration and equations where acceleration is not constant we are going to use the powerful method of calculus to look at these equations and we&amp;#39;ll also use the calculus method to derive the equations of motion I&amp;#39;m going to make these Concepts really easy for you and explain you with some simple examples so make sure you watch the video till the end and do check out the full courses on our website and Android app Manoa Academy I&amp;#39;ll put the links below and the best part is you can try out these courses for free so do check it out links are given below all right let&amp;#39;s dive into kinematic equations I&amp;#39;m sure you&amp;#39;ve seen these equations before what are these equations these are the equations of motion and remember the symbols here U is initial velocity V is final velocity T is time s is displacement and a is acceleration so these are equations of motion for an accelerated motion but there&amp;#39;s an important condition here the acceleration needs to be constant so the a that you see in these equations represents acceleration and that must be constant which means that the velocity should be increasing or decreasing at a constant rate but what if we have a motion where the acceleration is not constant what do you do then that&amp;#39;s exactly what we&amp;#39;re going to look in this video because we will look at the kinematic equations for Accelerated motion and we&amp;#39;ll use the mathematical concepts of calculus and look at these equations and I&amp;#39;m going to make these Concepts really clear for you so let&amp;#39;s dive right into it let&amp;#39;s start with the concept of velocity what is velocity remember velocity is defined as the rate of change of displacement and the formula is displacement divided by time so if a body has a certain displacement or a change of displacement you can denote that as Delta s over a certain time interval which we can denote as delta T so the change in displacement by change in time is the formula of velocity so let&amp;#39;s take an example let&amp;#39;s say a body travels a distance of 10 m so the displacement is 10 m here and let&amp;#39;s say the time interval is 5 seconds so we have me measured its displacement over 5 seconds so what will be the velocity of the body very simple change of displacement by time so 10 m by 5 Seconds it&amp;#39;s going to be 2 m/s now we have measured this over a 5sec time interval now within that time interval maybe the body was changing its velocity right maybe it was going a little faster a little slower just like when a car travels on the road it doesn&amp;#39;t perfectly travel at a constant speed right or at a constant velocity so maybe it covered this 10 m in 5 seconds but the velocity was changing so what have we really measured that&amp;#39;s right we have measured the average velocity the average over this 5sec time interval so remember this important concept that this formula is really the average velocity over your time interval and the average turned out to be 2 m/s but now let&amp;#39;s say we want to find the speed at a particular instant like in the cars you have seen there&amp;#39;s a speedometer right so it shows the speed of the vehicle at that instant not over an average time because whenever you&amp;#39;re looking at this uh speedometer let&amp;#39;s say the car is moving at a very high speed right or your bike is moving at a high speed you will see it shows 60 km/ hour 80 km/ hour when you are stopped it will show zero when you&amp;#39;re going slow 10 km/ hour so you have seen the speedometer constantly changes and it is giving you the magnitude of this velocity the speed at that instant so that is what we call instantaneous velocity and how will we get that to get the instantaneous velocity we must make this time interval very small we cannot meas measure it over a long time interval otherwise that will become an average you need to measure it over very short interval like1 second or 01 second or 01 second make that time interval very small so in physics or mathematics we write it as limit delta T tending to zero so when this time interval tends to zero that is denoted as DS by DT so here we are using the mathematical tools of calculus here where the short time interval we denote it as DT so DS by DT is basically your instantaneous velocity where s is displacement and T is time so you&amp;#39;re doing a differentiation with respect to time of the displacement so remember this thing that your instantaneous velocity your instantaneous velocity is going to be DS by DT so let&amp;#39;s write down that important formula V = to DS by DT and you have done graph stuff also right so we can say that if you draw draw a graph of displacement versus time velocity is basically the slope of that graph because when you do DS by DT or delt s by delta T you&amp;#39;re finding out the slope remember slope was y by X or change in y by change in X so basically velocity is the slope of the displacement time graph and you can calculate it as tan Theta at a particular point but we&amp;#39;ll talk about slope Concepts in a separate video in this video let&amp;#39;s focus on these equations and calculus stuff to understand the concept and formula let&amp;#39;s look at an example so here we&amp;#39;ve been given that displacement of a moving body is given by this equation S = 5 T ^2 + 4T so we can know the relation between displacement and time and what do we need to find here we need to find the instantaneous velocity at time T = 2 second and also the average velocity so let&amp;#39;s first start off with the instantaneous velocity so guys what is the instantaneous velocity formula remember V = to DS by DT so we are going to use this formula V = DS by DT for instantaneous velocity always use DS by DT Delta s by delta T is for average velocity right so here I can get the instantaneous by substituting s over here so we will do DDT that is a differentiation of the displacement which is 5T sare + 40 and you guys know how to do this this is all calculus stuff so please revise your Calculus because it is very very important and very useful for this chapter so we have to do a difference of this displacement and we know that we can do differentiation of each of the terms separately so this is basically going to be become DDT of 5 t² right plus DDT I&amp;#39;ll break it down of 4T now you guys know how to do the differentiation you bring the constant outside so this is finally going to become 5 DDT of t² plus 4 * DDT of T now what is DDT of t² in calculus you know this formula that DD X of x ^ n is n * x ^ n -1 so here my variable is not X it&amp;#39;s t so let&amp;#39;s apply that so DDT of t² will be become n * T ^ N - 1 here n is 2 so 2 * T ^ 2 - 1 which is 1 so 2 * T so use that formula and you will basically get 5 * 2T here plus 4 * what is DDT of T that&amp;#39;s simple it&amp;#39;s one again you can apply that formula it will be 1 * T ^ 1 - 1 which is T ^ 0 so again 1 so this is going to be + 4 so here we have got our instantaneous velocity V is 5 * 2 so that is 10t + 4 so that is my expression for instantaneous velocity so that is the general expression in terms of T but the question said find the velocity at time t equal to 2 second so it&amp;#39;s very easy all you have to do is substitute tal to 2 second here done so what is the velocity at 2 seconds we can say at T = 2 second the velocity is going to be V = 10 * 2 + 4 so what is the answer 20 + 4 24 and guys don&amp;#39;t forget the units 24 m/s actually here we should have been given the units and meters you know so that is missing in the question here that the displacement should have been given in meters since time is given in seconds uh but if the units are not given then you&amp;#39;ll have to write 24 units okay whatever the unit is kilometer per hour meter per second so there you can see that is our final answer here so that&amp;#39;s all you have to do if they ask you instantaneous velocity do the differentiation do DS by DT and if they ask you at a particular time just substitute it and you&amp;#39;re done next let&amp;#39;s look at the second part which is average velocity between 2 seconds and 3 seconds now let&amp;#39;s calculate the average velocity it&amp;#39;s been asked to find the average velocity between time 2 seconds and 3 seconds so for that we&amp;#39;re going to use the average velocity formula so my average velocity formula is Delta s by delta T change in displacement by change in time so we can write that as S2 - S1 the change in displacement divide by change in time T2 minus T1 now what is the time here T1 is 2 seconds and T2 is 3 seconds right that is the time which has been given here and what is the displacement at these times the displacement we can get using this formula S = 5 t^2 + 4T you just have to substitute the time over there very simple so let&amp;#39;s work it out so S2 which means the displacement at 3 seconds is going to be 5 * 3 2 + 4 * 3 minus the displacement at 2 seconds so just substitute T = 2 there so 5 * 2 2 + 4 * 2 so that&amp;#39;s all we&amp;#39;ve just substituted the time values in the displacement expression div divided by the t2 - T1 3 - 2 so simple there you go so now you just have to do your calculations here so let&amp;#39;s go ahead and do that what will this be 5 * 3 2 that means 5 * 9 this is going to be 45 + 4 * 3 is 12 minus uh this we need to be careful about this entire thing should be in minus right in uh should be under the brackets over here because we are doing S2 - S1 so 5 * 2 2 is 5 * 4 so that&amp;#39;s going to be - 20 and we have minus 4 * 2 is 8 over there divided 3 - 2 is 1 so just calculate what will you get over here you&amp;#39;re getting 45 + 12 so which is basically 57 - 28 / 1 so the answer is 29 m/s this is our average velocity average velocity 29 m per second and that is from the time 2 to 3 seconds so see it&amp;#39;s very simple just use the formula and you can get the average velocity and just carefully substitute the values calculate and there you get your answer so we have seen how to find velocity from displacement now how do you find displacement from velocity so we know velocity is displacement by time so you can rearrange the equation and you&amp;#39;ll get displacement equals velocity multiplied by time so simple but this equation is representing your average velocity so if you know your average velocity you multiply by the time you will get the total displacement in that time interval right so total displacement is average velocity multiplied by time but what if the velocity of the body is changing over the time and you don&amp;#39;t know the average velocity so let&amp;#39;s say we want to work with the instantaneous velocity then what do we do so remember for instantaneous velocity how do we calculate it from the displacement the formula was the instantaneous velocity so I&amp;#39;ll write I N St that is for instantaneous velocity equals DS by DT so to calculate it what do we do we do a differentiation with time we do DS by DT so you can see here we are doing a difference here we are doing a differentiation this DS by DT is the differentiation in calculus and that way you can find out the instantaneous velocity the velocity at a particular instant now suppose we know the velocity at certain instances right we know the instantaneous velocity and I want to find the displacement from that so how do I do that so let&amp;#39;s see if we can use this formula so if you look at this formula I&amp;#39;m interested in calculating displacement so I can rearrange this and write it as if I bring the DT here so we can do V * DT = to DS since we interested in displacement let&amp;#39;s bring that to the left hand side so or I can write DS = to vdt what is the meaning of this equation that in a very small time interval if you multiply the instantaneous velocity you will get that small instantaneous displacement in that small time interval but we want to find the displacement over a certain time so what do you have to do we have to add up all of these so vdt so the vdt values need to be added up across that time interval and how do you do that so here again we use the calculus method of integration because finding that area right adding up all these vdts finding the summation here of all this is what we represent in calculus as integration because we want to find the total displacement so if we integrate both sides if we do integration of DS equal to integration of vdt what will we get therefore we get S because when you integrate displacement you get displacement it will be the integration of vdt so you can see so you can see it is exactly the reverse process to get instantaneous velocity from displacement we do differentiation V equal to DS to get displacement from velocity naturally you&amp;#39;ll do the opposite thing so instead of differentiation you are going to do integration here and integration of vdt over the time interval will give you displacement now you might be thinking what is the time interval that will be given to you so normally you&amp;#39;ll see in the books they say over a certain time interval 0 to T or you can write T1 to T2 so that will end up giving you the displacement so remember this very important concept and here again we are talking about instantaneous velocity right and the total displacement calculated from that that is why we are using all this calculus stuff dsdt and this integration of vdt don&amp;#39;t get scared by this calculus terminology I know initially it sounds confusing and scary but it is just a transformation of the basic concept that you have learned we just applying calculus here and we are getting these formulas clear so let&amp;#39;s understand how we can calculate displacement from instantaneous velocity using this example so here we have been given that velocity of a moving body is given by V = 10t we need to find the displacement from 0 to 4 seconds so here you can see the velocity of the body is constantly changing because as the time changes it is 10 * T okay so at time equal to 0 it is 10 * 0 0 at time let&amp;#39;s say at 1 second it becomes 10 m/s at 2 second it becomes 20 m/s at 2.5 it will become 25 m/s and so on so clearly you can see velocity of the body is constantly changing and this is the instantaneous velocity this gives us the formula of the Velocity at any time T so how do we find the displacement from 0 to 4 seconds so here once again we have to use that integration concept because we have been given instantaneous velocity and what was our formula to find the displacement from instantaneous velocity you need to integrate the velocity over the time interval so we have to integrate V DT okay and the time interval here has been given from 0 to 4 seconds so put the limits from time 0 to 4 seconds this is the integration and that will give me the total displacement that&amp;#39;s it so go ahead and substitute this 10t into the formula so what will we get from integration from 0 to 4 seconds and this is 10t the velocity is 10t DT now how do you do this integration whenever there&amp;#39;s a constant you bring it out so this will be integration from 0 to 4 seconds the 10 comes out T DT now what is the integration formula and calculus when we do integration it is integration of x ^ n DX is going to be x ^ n + 1 / n + 1 this is the formula integration is the opposite of differentiation there you had n minus 1 here there is n + 1 right that was n * x ^ nus1 so here what is our power over here you can see this is T that means this is basically T to to the^ 1 so we just need to substitute n = 1 over here so basically what happens so this will give us 10 times and integration of T DT will be T ^ 1 + 1 t² / by 2 so we are getting t² by 2 and the limits here are from 4 to 0 so the limits for this are from 4 to 0 so what does that mean that I need to substitute four here and 0o and then subtract them right because when you have limits you basically subtract here so it&amp;#39;s going to be 10 * T ^2 by 2 so with the value 4 here so that&amp;#39;s going to be 4 2 by 2 and then you substitute Min - 0 2 by 2 right and so what do we end up getting this will become 10 and this 4 S by 2 is basically 16 by2 that&amp;#39;s 8 8 - 0 and this is going to be 8 * 10 which is 80 M 80 m is your answer again uh this should have been given all in SI units so it should be mentioned here you know these are all in SI units and your answer is going to be 80 M for the displacement so remember this important way of doing integration and finding the displacement and also you guys should know that when you are doing this integration when the limits are not given it is called an indefinite integral then we put a constant over there plus C but when the limits are given then this constant is not there so that is why I did not add the constant here because the limits were given to us from 0 to 4 seconds if that was not given then you need to add a plus constant that is indefinite integration indefinite integral and this was a definite integral because the limits were given to you we substituted that and there you can see we have got our answer now let&amp;#39;s accelerate to acceleration what is acceleration the change in velocity divided by time because it is defined as the rate of change of velocity so the change in velocity by the change in time we can also represent that in a formula like this acceleration a is Delta V that is the change in velocity divided by the time interval delta T and you&amp;#39;ve all seen this formula right Delta V if you represent as final velocity minus initial velocity so vus U / T so let&amp;#39;s say the time is from 0 to T but now what if the acceleration is changing right what is is the acceleration at a particular instant just like we had instantaneous velocity how can we Define instantaneous acceleration because once again this will give me the average acceleration over a large time interval so what this is really is really your average acceleration because acceleration need not be perfectly constant all the time that will become a case of uniform acceleration so how do I Define that or what is the formula going to be so again if you want to measure the the instantaneous acceleration in a very very very small interval of time what do we need to do again we&amp;#39;ll use calculus we&amp;#39;ll say in the limit delta T tending to zero that means a very small time interval my instantaneous acceleration instantaneous this is a very interesting word instantaneous it has all the vowels see a e i o U right in fact I think I just noticed that while writing I hope I got my spelling right it&amp;#39;s a long word so instantaneous acceleration is going to be a will be when the limit tends to zero we denote it as DV by DT this means very small time interval so acceleration is DV by DT this is the formula for instantaneous acceleration and once again you&amp;#39;ll have that slope concept right so if you&amp;#39;re having a velocity time graph acceleration is DV by DT so it is basically the slope of the Velocity time graph so remember these formulas this is the formula for average acceleration and instantaneous acceleration you&amp;#39;re doing a differentiation of the Velocity with respect to time best way to understand formulas is with the help of an example so let&amp;#39;s take a look at this example here we have been given displacement of a moving body s = 5 t^2 + 4T and we need to find the instantaneous acceleration of this body so how will you do it remember the instantaneous acceleration formula instantaneous acceleration a is DV by DT we do a differentiation of the Velocity but what has been given here is displacement we have been given S no problem we are going to first find velocity and then plug it in here so we can start off with this relation V equal to the instantaneous velocity V equal to DS by DT so let&amp;#39;s plug in this uh displacement over here so we have DDT and the displacement is 5 t² + 4T so go ahead and do a differentiation of that so if you do DDT of 5 t² it&amp;#39;s going to be 5 DDT of t² + 4 * DDT of T so what will you end up getting DDT of t^2 like we saw is 2 * T you can use that n * x ^ nus1 formula so this will give you 5 * 2T and DDT of T is 1 you can easily remember that like this right DD DT and DT seems to be cancelling so that&amp;#39;s going to be + 4 so I&amp;#39;m getting the 10t + 4 in fact we had seen this earlier when we had calculated the instantaneous velocity now we need to plug this in here so what will we get the instantaneous acceleration will be DDT and we are basically plugging in the Velocity in this formula so our velocity is 10 t + 4 the instantaneous velocity so do a differentiation here what will you get again 10 * DDT of t plus DDT of 4 four is a constant so its differentiation will be zero so therefore finally we are getting DDT of T is 1 so 10 is the answer acceleration is 10 if units are not given you can simply write 10 units if these are given in seconds and all you know or SI units you know it&amp;#39;s going to be m/s squar so here in this case the instantaneous acceleration value turned out to be constant so it is actually a case of uniform acceleration but it need not be maybe you could have got an expression like 10t or you know 5T or something like that so this is what we have got here so basically you can see you can find acceleration as differentiation of the Velocity if displacement is given it&amp;#39;s like a double differentiation so if you think about it a is DV DT and if we substitute v as dsdt so this is like a double differentiation so basically what you&amp;#39;re doing a DDT squ of your displacement so this is another way to look at acceleration or you can start off with this and do the steps and get it and remember here they just said instantaneous acceleration they didn&amp;#39;t ask you to find acceleration at a particular time so when they give a question like this you can leave it in terms of T suppose we got 10t over here you leave it as that because they have not given you that I want to know the acceleration at this time at 4 seconds 5 Seconds nothing like that is given so you can leave the expression in terms of T like that we have seen how to calculate acceleration from velocity but how do you calculate velocity from acceleration so of course you can rearrange the simple formula and you will get velocity is acceleration multiplied by time time but the Restriction here is that acceleration should be constant so for uniform acceleration this works great you can get the velocity but what if the acceleration is changing like in the case of instantaneous acceleration so once again we have to use the concept of calculus remember how did we Define instantaneous acceleration instantaneous acceleration is the differentiation of velocity so DV DT so to find acceleration from velocity what were you doing differentiation with respect to time so we were using differentiation so I think you must have guessed that how do you calculate velocity from acceleration you have to do the opposite of differentiation which is basically integration and that you can easily see from this equation because if you rearrange it so if you multiply the take the time this way so you get a DT equal to DV since I want to find velocity I&amp;#39;ll bring it on the left hand side so I can say DV equal to a * DT what does this mean that if you multiply acceleration with a very very small time interval you will get the value of that small velocity and to get the full velocity you need to take a summation here right summation in statistics we represent with Sigma but in calculus we write that big kind of of that kind of s symbol for integration so what you really need to do is if you want to calculate velocity from acceleration you need to integrate both sides we can integrate both sides of the equation so on integrating both sides integral of DV will give you the velocity that&amp;#39;s our goal here and this will be the integration of acceleration time DT again remember if uh the time interval is not given then obviously you it&amp;#39;s a indefinite integral you have to take plus a constant but if the time interval is uh given then you directly substitute the values here so let&amp;#39;s say you want to calculate the velocity from for the time interval 0 to T so your integration formula is going to look like this so remember this important concept that to get instantaneous acceleration from instantaneous velocity you do DV by DT but if you want to get the velocity the instantaneous velocity then you have to do integration of the acceleration so you have to do exactly the opposite and to get velocity from acceleration this time you need to do integration so very very important Concepts please understand it and write it down so let&amp;#39;s see how we can calculate velocity from acceleration using this example so what have we been given here acceleration of a moving body is given by a = 3T ² and we need to find the velocity at Time 4 seconds if initially the body is at rest so some condition has also been given to us so once again let&amp;#39;s go ahead and apply our formulas of calculus and see what we can get here so our goal is to find velocity from acceleration the instantaneous acceleration is given to us a equal to 3 t² acceleration is dvdt but velocity is going to be the integration the opposite right so if I want to calculate velocity it is going to be the integration of the a DT right and if a time interval is not given then it&amp;#39;s an indefinite integral like this so let&amp;#39;s go ahead and substitute the acceleration expression here it&amp;#39;s going to be integration of 3T Square DT so what will that be guys you bring the three outside it&amp;#39;s a constant integration of t² DT and once again here&amp;#39;s use your maths formula that integration of x ^ n DX is going to be x ^ n + 1 by n + 1 and if it&amp;#39;s indefinite you will have a constant over there plus C so since no time interval is given I&amp;#39;ll I have to take that constant plus C here so what will be integration of t² T Cub by 3 right so this is going to give me 3 * T Cub by 3 plus a constant C and this three and three will cancel here so what are we left with TQ + C so velocity is going to be equal to TQ + C this is my expression and I need to find the velocity at tal to 4 second so if I simply substitute four over here I&amp;#39;m still left with this plus C this constant is irritating over here right I don&amp;#39;t know its value but we can find out its value because we have been given a condition that initially the body is at rest initially means time equal to zero so for this special condition given to us at time equal to0 we know that velocity is zero so see using this can you calculate the constant of course you can so if you substitute it here what will you get put V = to 0 t = to 0 so 0 CU plus C so therefore you&amp;#39;re getting your constant C is Zer so that&amp;#39;s it we know the constant is zero so finally we can get velocity at 4 second so at T = 4 second V = TQ Q which means 4 CU which means 64 again units right or if the it&amp;#39;s in SI units then it will be 64 m/s so here we are all assuming SI units here so your final answer is going to be 64 m/ second 4 Cub so see just apply your integration and again remember if limits are not given then you need to take a plus constant if limits were given to you then you&amp;#39;ll have to just simply subtract suppose the limits were given let&amp;#39;s say 2 to 3 then you just have to do 2 uh TQ uh from the value at 3 minus the TQ value at 2 so it will become 3 Cub minus t CU But Here No Limits were given so we took a plus c and easily calculated our answer here here are the three famous equations of motion and as we discussed what is the restriction here the acceleration a is constant so these are all uniform acceleration where the acceleration is a constant value now you must have studied how to derive these equations of motion using algebraic methods and using graphical methods but here we&amp;#39;re going to do something very interesting we are going to derive it using calculus so we will start with those instantaneous velocities or uh instantaneous acceleration all those Concepts that we learned with all this differentiation integration and we are going to apply all those equations to derive these equations because remember the equations that we&amp;#39;ve been looking at using calculus is for any general motion right the velocity does not have to be constant the acceleration does not have been to be constant because it deals with instantaneous velocity instantaneous acceleration so those equations are valid in the general case and let&amp;#39;s see if we can apply that to get our special equations of motion where acceleration is constant so we&amp;#39;ll start with the derivation of the first equation of motion here let&amp;#39;s look at the derivation of the first equation of motion v = u + a this time using calculus technique so we will start with the equation of acceleration the calculus equation for instantaneous acceleration we know a = DV by DT and the main trick about these derivations is knowing the starting step where to start after that you&amp;#39;ll see the flow is easy so we know acceleration is dvdt now in this equation you can see the goal is to write velocity the final velocity V is initial velocity U Plus 8 right we want to uh derive this equation so clearly we want to find V in terms of a so what do we do we&amp;#39;ll rearrange this equation as we&amp;#39;ve done before and so we&amp;#39;ll write DV equal to a DT and we know that if we integrate both sides we should get the velocity so I can apply integration on both sides right now what will be the limits of integration here in this equation you know we take the body is going from a time interval 0 to T so we can see that the integration for the time will be done in the limits 0 to T and the velocity at Time Zero we denote with u that is the initial velocity so here the limit will be U and at time T we call it V so you can see those are the limits for our velocity so now go ahead and do the integration and see if you&amp;#39;re able to get that equation so when we do the integration on the left hand side we have DV so integration of DV is going to be V and then you apply the limits here from U to V initial to final velocity what do we have on the right hand side on the right hand side we have integration a DT but remember in this the acceleration is constant that is the Restriction of the equations of motion it is a constant acceleration the rate of change of velocity is constant velocity is not constant the rate of change of velocity which is the acceleration is constant since it is a constant term I can bring it outside right of the integration so what will I get a comes out and we have integration 0 to T of DT so on the right hand side we&amp;#39;ll have basically integration of T will give you integration of DT will give you T and we apply the limits here from T to zero and here you have V with the limits V to U now what is the meaning of this limits basically you set this variable equal to V and U and take the difference so it is basically left hand side becomes V minus U you substitute V and minus the lower limit upper limit minus the lower limit same way over here upper limit minus the lower limit you substitute into T so it&amp;#39;s going to be a * T minus 0 now go ahead and simplify this this is very easy v- u = a t and if you rearrange it what do you get v = u + a and there you go we have our first equation of motion and yes we have derived it using calculus so we just applied simple integration here starting with the uh formula of acceleration or you can start off with DV equal to a DT right and there we have got our first equation of motion now let&amp;#39;s take a look at the second equation of motion now let&amp;#39;s look at the derivation of the second equation of motion which is s = UT + a² remember s is displacement U is initial velocity T is time and a is acceleration so we can see we are clearly interested here in displacement there&amp;#39;s velocity time acceleration so let&amp;#39;s start off with the velocity formula V = dsdt since we want to calculate displacement let&amp;#39;s bring that on the left hand side so we&amp;#39;ll bring velocity and time on the other side because you can see in the equation of motion displacement is on the left hand side here so I can write DS equal to vdt right DS equal to V DT now here we could do an integration but you can see that the second equation of motion doesn&amp;#39;t have V the final velocity it has U the initial velocity so this is where you can apply your first equation of motion V equal to U + a and change this in terms of U and a so what you&amp;#39;re going to do is substitute V = to U + 8 from the first equation of motion we are allowed to use the first equation because we have proved it okay so we are getting DS equal to U + a * DT now I want to find the displacement that means we need to do an integration of this expression because I&amp;#39;ve got DS the small displacement we need to add it up so we will do an integration on both sides and this will get me the displacement from 0 to S and here the limits will be 0 to T So once again go and apply your Calculus fundas and do this integration so what will you get here if you integrate DS on the left hand side we&amp;#39;ll get S with the limits s and Zer okay and what do we get on the right hand side when we are integrating this so we will get this expression is U + a * DT so when we are integrating it we will get integration of from 0 to T of U DT plus integration of 0 to T of a a t * DT you can break the plus down like this so let&amp;#39;s go ahead and do that integration here initial velocity is a constant number so we&amp;#39;ll bring it out so this becomes U integration from 0 to T of DT acceleration is constant in the equations of motion so we can bring that out acceleration comes out here integration from 0 to T of T DT so what will you get over here integration of DT is T and you have the limits T to zero and here we have a integration of T DT is going to be T ^2 / 2 so what we have over here T ^2 by 2 again the limits of T and z and on the left hand side we have S with the limits s and Z so go ahead and substitute the limits and you&amp;#39;re done so the left hand side basically becomes s- 0 which is s or you can write s - 0 here this becomes u u t minus 0 is t plus half a t² since again it is from T to 0o so we&amp;#39;ll put t² here minus 0 will of course give you Zer so there you see we are getting s = u + half a [Music] t² and we are done we have derived the second equation of motion using our calculus method again the main trick is to know how to start and you go ahead and do the calculus method you apply your integration and you will easily get the derivation here now let&amp;#39;s look at the derivation of the third equation of motion by calculus method the third equation of motion is v² = U ^2 + 2 a s we have acceleration a in here so using calculus method we can first start with the formula of acceleration what is the formula of instantaneous ACC acceleration instantaneous acceleration a is DV by DT but if you just try to integrate this equation there&amp;#39;ll be a problem because we don&amp;#39;t want the time variable there&amp;#39;s T in here but the third equation of motion does not have any time so we have to do some magic and get rid of this time and what we really need is a v they are fine and we need displacement s so what can we do here let&amp;#39;s take a look so what you can do you can change this into to get rid of time you do DV by DS and since we are dividing by DS I&amp;#39;ll again multiply by DS so that it cancels and I&amp;#39;ll shift the DT here so write down this is the magic step to get rid of the time variable how do we get rid of the time because what is DS by DT s is displacement so DS by DT is rate of change of displacement that means it is velocity so this part so this term here is simply velocity V so what can we write this as we can write acceleration a is DV by DS multiplied by the velocity V here so see time has been kicked out time is not there now what do we do just cross multiply bring the DS this side and let&amp;#39;s do some maths integration so what will we get here so we can can do a DS equal to V DV okay and now to get that equation let&amp;#39;s integrate on both sides so when you integrate on both sides so what will be the integration limits here so let&amp;#39;s say initially the displacement is zero at the initial point right and the final displacement is s and when I integrate velocity the initial velocity we denote by U and Final velocity by V so these are my integration limits now go ahead and do the integration for these equations of motion you know the important thing that acceleration is constant so since a is constant we can bring it outside the integration so what will this become a and integration of Simply DS here we have integration of V DV so inte ation of V DV is going to be v² by 2 using that integration formula and the limits will be from U to V and when you do integration of this what will you get you will simply get S so integral of DS will be S and the limits will be from 0 to S so now just plug in the limits and what do we get a * s - 0 equals we&amp;#39;ll plug in these limits here v² - U ² or v² by 2 - u s by 2 so you can have two in your LCM now just cross multiply and see what you get so if you bring the two over here we are going to get 2 a s on the left hand side and right hand side will be v² - U ² you can see we have got to the third equation of motion just rearrange it and and you will get v² = U2 + 2 a s there you go we have derived the third equation of motion using calculus method the main trick again is how to start we started with the acceleration formula and this was the magic trick you have learned in fact you can see there&amp;#39;s a sort of a new formula you have here a = v DV by DS or a = d DV by DS * V and we use this to integrate and got our third equation of motion so simple here&amp;#39;s our formula summary that will help you revise what we have learned so remember we started with these three equations of motion 1 2 and 3 and you have done these before in class N9 right V = to U + a s = to UT +/ a s and v² = to U2 + 2 a s these are the three famous equations of motion but they have a condition what is the condition even though they are accelerated motion the acceleration should be constant so here the a the acceleration in these equations of motion is constant it is a fixed value for example 5 m/s Square 10 m/s square whatever be the value it is fixed so it&amp;#39;s accelerated motion where velocity is changing but acceleration is fixed then you can use these equations but if acceleration is changing right or we have this instantaneous velocity instantaneous acceleration then what do we do then please don&amp;#39;t use these equations you need to use these equations these are the equations for this instantaneous velocity instantaneous acceleration so I&amp;#39;m going to write that word over here instantaneous so that you can instantly remember what to use and what is the instantaneous formula V equal to DS by DT remember average velocity was Delta V by delta T So instantaneous velocity is DS by DT similarly acceleration was Delta V by delta T instantaneous acceleration is DV by DT so what are we doing we are doing a differentiation here with respect to time we using calculus so you can get velocity from displacement by doing differentiation how can we get the displacement from the velocity how do we do that you just do the reverse you do integration reverse of differentiation is integration so the formula is s equal to integration of v d v DT it&amp;#39;s easy to remember just take the DT this side and integrate same way just like you can get acceleration from the velocity by doing differentiation you can get velocity from acceleration by doing integration just the reverse so I hope this formula summary will help you remember these important Concepts and do write down these formulas the best way to remember formulas is to write them down not just look at them okay so go ahead and jot them down and I&amp;#39;m sure you&amp;#39;ll remember them now I have an interesting question for you the question says displacement of a moving body along the x- axis is given by x = 60 + T Cub so the displacement is given in terms of time you need to find the acceleration at time equal to 5 seconds so go ahead and try this question and let me know your Answers by putting it in the comments below and I promise to reply to your answers so go ahead apply the concepts that you&amp;#39;ve learned and try out this question and let me know your answers by putting it in the comments below friends I hope these kinematic equations of accelerated motion are crystal clear to you now and remember practice makes you perfect so go ahead and apply these Concepts and practice more questions on this topic and if you like this video hit the like like button and share it out with your friends and if you haven&amp;#39;t subscribed to our YouTube channel come on hit that subscribe button and click on the notification Bell so that you don&amp;#39;t miss out on any of our videos and do check out the full courses on our website and Android app I&amp;#39;ll put the links below and the best 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Transcript for:Kinematic Equations of Accelerated Motion

Transcript for:
Kinematic Equations of Accelerated Motion