Finding Position from Velocity or Acceleration

Question

Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.

v = sin πt, s(0) = 0

Accepted Answer

Welcome back, everyone. In this problem, the velocity of an object moving along a straight line is defined by V equals the cosine of pi T. Its initial position is S 0 equals 2. Using the second corollary of the mean value theorem, determine the object's position at a time T. Now, how is the 2nd corollary of the mean value theorem going to help us to determine our object's position? We will recall that this tells us that if the derivative of our function F of X is equal to the derivative of our function G of another function G of X, OK, at each point. X in an open interval AB. Then there exists a constant sea. OK, such that. F of X will be equal to G of X, A, plus C for all values of X in our interval AB, OK? No, we know our problem here. What it's basically telling us is that F minus G is going to be a constant function on AB, the open interval AB. So if we can find the derivative of our position function, OK, then we should be able to find the derivative of this other function that satisfies the condition, and we can use it to help us solve for our constant C and thus our function F of X. Now, let's, let's localize it for the information we're looking for, OK? We want to find the object's position at that time T. So what we're really saying for our problem then is that if the derivative of SFT, OK, is equal to the derivative of some other function G of T, OK. At each point T in or interval AB, then there is a constant C such that, OK. And, and now, let's just write this out here. Such that S of T. Will be equal to G of T plus C for all values of T K, that's in or open interval AB. Now what do we know about the derivative of SFT, the derivative of our position function? Well, We know that the velocity V is the derivative of the positions with respect to time T, OK? And our problem tells us that our velocity is the cosine of pi T. So this would imply. That G of T is equal to. Let's write that out here. G T. The derivative of GF T would also be equal to the cosine of pi T. And no, the question we're asking ourselves is what can we differentiate to get the cosign of IT? Well, if we think about it, G of T must then be equal to 1 divided by pi multiplied by the sine of pi T. Because when we differentiate this, then it would be equal to pi multiplied by 1 divided by pi cosine of pit which would give us the cosine of pi T, OK. So no, since they both have the same derivative. Then this would imply that S of T is going to be equal to 1 divided by pi multiplied by the sign of pi T plus C. How can we solve foresee here? Well, again, remember we were told in our problem statement that the initial position of our object is S of 0 equals 2. OK? So now, based on what we know what our function at T equals 0, S would be equal to 2, and thus we can use that to solve our constant because the constant is going to be the same of course it remains constant regardless of of whatever point. So no. That means that S of T2 is going to be equal to 1 divided by pi multiplied by the sign of pi multiplied by T0 plus C. If we think about it, the sign of pi multiplied by 0 is 0. The sign of 0 equals 0. So thus this must mean that C equals 2, and if C equals 2, then that means our functions of T must be equal to 1 divided by pipit. Plus 2. In other words, The function that represents the position of the object with a velocity of a cosine of pi T and an initial positions of 0 equals 2 is going to be S T equals 1 divided by pi sine pi T + 2. Thanks a lot for watching everyone. I hope this video helped.

Finding Position from Velocity or Acceleration

Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.

v = sin πt, s(0) = 0

Key Concepts

Integration

Initial Conditions

Fundamental Theorem of Calculus

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