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# Worked example: motion problems (with definite integrals)

AP.CALC:
CHA‑4 (EU)
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CHA‑4.C (LO)
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CHA‑4.C.1 (EK)

## Video transcript

let's review a little bit of what we learned in differential calculus let's say we have some function s that it gives us as a function of time the position of a particle in one dimension if we were to take the derivative with respect to time so if we were to take the derivative with respect to time of this function s what are we going to get well we're going to get DS DT or the rate at which position changes with respect to time and what's another word for that the rate at which position changes with respect to time well that's just velocity so that we could write it as velocity as a function of time now what if we were to take the derivative of that with respect to time so we could either view this as the second derivative we're taking the derivative not once but twice of our position function or you could say that we're taking the derivative with respect to time with our velocity function well this is going to be we can write this as we can write this as D V DT the rate at which velocity is changing with respect to time and what's another word for that well that's also called acceleration this is going to be our acceleration as a function of time so you start with the position function take its of the position as a function of time take its derivative with respect to time you get velocity take that derivative with respect to time you get acceleration well you could go the other way around if you started with acceleration if you started with acceleration and you were to take the antiderivative if you were to take the antiderivative of it the aunt D an antiderivative of it is going to be actually let me just write it this way so an antiderivative I'll just use the integral symbol to show that I'm taking the antiderivative is going to be the integral of the antiderivative of a of T and this is going to give you some expression with a plus C and we could say well that's a general form of our velocity function this is going to be equal to our velocity function and to find the particular velocity function we would have to know what the velocity is at a particular time then we could solve for our C but then if we're able to do that and we were to take the antiderivative again then now we're taking the antiderivative of our velocity function which would give us some expression as a function of T and then some other constant and if we could solve for that constant then we know then we know what the position is going to be the position is a function of time just like this we would have some plus C here but if we know our position at a given time we could solve for that C so now that we've reviewed it a little bit but we've rewritten it in I guess you could say thinking of it right not just from the differential point of view for the derivative point of view but thinking of it from the antiderivative point of view let's see if we can solve an interesting problem let's say that we know that the acceleration of a particle is a function of time is equal to 1 so it's always accelerating at one unit per and you know I'm not giving you time let's let's just say that we're thinking in terms of meters and seconds so this is one meter per second one meter per second squared right over here that's our acceleration as a function of time and let's say we don't know the velocity expressions but we know the velocity to a particular time and we don't know the position expressions but we know the position at a particular time so let's say we know that the velocity at time three let's say three seconds is negative three meters per second and actually want to write the unit's here just to make it a little bit a little bit so this is meters per second squared that's going to be a unit for acceleration this is our unit for velocity and let's say that we know let let's say that we know that the position at time two at two seconds is equal to negative 10 meters so if we're thinking in one dimension of which is moving along the number line it's ten to the left of the origin so given this information right over here and everything that I wrote up here can we figure out the actual expressions for velocity as a function of time so not just velocity at time three but velocity generally as a function of time and position as a function of time and I encourage you to pause this video right now and try to figure it out on your so let's just work through this what is we know that velocity as a function of time is going to be the anti derivative the anti derivative of our acceleration is a function of time our acceleration is just one so this is going to be the antiderivative of this right over here is going to be T and then we can't forget our constant plus C and now we can solve for C because we know V of three is negative three so let's just write that down so V of three is going to be equal to three three plus C I just so every place where I saw the the T or ever place where I have the T I replace it with this three right over here actually let me make it a little bit clearer so V of three V of 3 is equal to three plus C and they tell us that that's equal to negative three so that is equal to that is equal to negative three so what's C going to be so if we just look at this part of the equation or just this equation right over here if you subtract 3 from both sides you get you get C is equal to negative 6 and so now we know the exact we know the exact expression that defines velocity as a function of time V of T V of T is equal to T T plus negative 6 or t minus 6 and we can verify that the derivative of this with respect to time is just 1 and when time is equal to 3 3 minus 6 is indeed negative 3 so we've been able to figure out velocity as a function of time so now let's do a similar thing to figure out position as a function of time we know that position is going to be an antiderivative of the velocity function so let's write that down so position as a function of time is going to be equal to the antiderivative of V of T DT which is equal to the antiderivative of t minus 6 DT which is equal to well the antiderivative of T is T squared over 2 so T squared over 2 we've seen that before the antiderivative of negative 6 is negative 6t and of course we can't forget our constant so plus plus C so this is what s of T is equal to s of T is equal to all of this business right over here and now we can try to solve for our constant and we do that using this information right over here at at 2 seconds where our position is negative 10 meters so s of 2 or I could just write it this way let me write it this way s of 2 at 2 seconds is going to be equal to 2 squared over 2 that is let's see that's 4 over 2 that's going to be 2 minus 6 times 2 so minus 12 plus C is equal to negative 10 is equal to negative 10 so let's see we get 2 minus 12 is negative is negative 10 plus C is equal to negative 10 so you add 10 to both sides you get C in this case is equal to 0 so we figured out what our position function is as well the C right over here is just going to be 0 so our position as a function of time is equal to T squared over 2 minus 6t and you can verify when T is equal to 2 2 squared over 2 is 2 minus 12 is negative 10 you take the derivative here you get T minus 6 and you can see and we already verify that V of 3 is negative 3 and you take the derivative here you get a of T just like that anyway hopefully you found this enjoyable
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