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Studying for a test? Prepare with these 8 lessons on Applications of definite integrals.
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# Worked example: motion problems (with definite integrals)

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 in which position changes with respect to time. And what's another word for that, position changes with respect to time? Well, that's just velocity. So that we can write 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 of our velocity function. Well this is going to be, we can write this as, we can write this as dv, 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 it, the position as a function of time. Take its derivative with respected time, you get velocity. Take that derivative with respected 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 anti, anti, an antiderivative of it is going to be, actually let me just write it this way. So an antiderivative, I'll just use the interval symbol to show that I'm taking the antiderivative. Is going to be the integral of the anti-derivative 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. And then, we could solve for our c. Whether then, if we're able to do that and we were to take the anti-derivative again. Then, now we're taking the anti-derivative 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, it would have some, plus c here 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 not just from the differential point of view from the derivative point of view. But thinking of it from the anti-derivative point of view. Lets see if we can solve an interesting problem. Lets say that we know that the acceleration of a particle is a function of time is equal to one. So it's always accelerating at one unit per, and you know, I'm not giving you time. Let, 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 at 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 I wanna write the units here, just to make it a little, a little bit. So this is meters per second squared, that's going to be our unit for acceleration. This is our unit for velocity. And let's say that we know, let's say that we know that the position at time two, at two seconds is equal to negative ten meters. So, if we're thinking in one dimension, of if this is moving along the number line, this is 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 own. 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-derivitive of our acceleration is a function of time. Our acceleration is just one. So this is going to be the anti-derivitive 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 3 is negative 3. So lets just write that down. So v of 3 is going to be equal to 3, 3 plus c. I just replaced where I saw the, the t or every place where I have the t I replaced it with this 3 right over here. Actually let me make it a little bit clearer. So v of 3, v of 3 is equal to 3 plus c, and they tell us that that's equal to negative 3. So that is equal to, that is equal to negative 3. 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 one. And when time is equal to 3, time minus 6 is indeed negative 3. So we've been able to figure out velocity a s 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 gonna be an anti-derivative of the velocity function, so let's write that down. So, position, as a function of time, is going to be equal to the anti-derivative of v of t, dt. Which is equal to the anti-derivative of t minus 6, dt which is equal to well the anti-derivative of t, is t squared over 2. So, t squared over 2, we've seen that before. The anti-derivative of negative 6 is negative 6 t, 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 two seconds were at, our position is negative two meters. So s of 2, or I could just write it this way. Well, 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 + 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 equals 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 6 t. And you can verify. When t is equal to 2, t 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 verified 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.