Finding Position from Velocity or Acceleration

Question

Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.

a = 9.8, v(0) = −3, s(0) = 0

Accepted Answer

Welcome back, everyone. In this problem, the acceleration of an object moving along a straight line is A equals 9.8. Its initial position and velocity are S 0 equals 2 and V 0 equals -5 respectively. Using the second corollary of the mean value theorem, determine the object's position at a time T. Now how is the second corollary of the mean value theorem going to help us to find the object's position? What do we know? Well, recall. That it basically tells us that if the derivative of our function FF X is equal to the derivative of another function G of X at each point X. OK. In an open interval AB, OK. Then There exists a constant sea. Such that OK F of X will be equal to G of X plus C, OK, for all values of X in that open interval, OK? In other words, F minus G is a constant function on AB. Now here if we were to use that to find the object's position, then we'd need the derivative of the position which is the velocity, and we don't have a function for our velocity here. However, we know that the acceleration of our object is 9.8 and the acceleration is the derivative of the velocity. So if we could use the acceleration to figure out a function for the velocity and then use that velocity as the derivative to the position, then we should be able to get an, a, a formula for the object's position. So let's go ahead and do that. So first, let's try to find uh our velocity function using the acceleration, OK? No, by the mean value theorem then. That means the of tea. The velocity is going to be equal to G of T plus C, OK, where the derivative of V of T is going to be equal to the derivative of G of T. Now, in that case, OK, we know the derivative of G of T or, or rather the derivative that we have is the acceleration, 9.8. OK. And since our derivative is equal to 9.8, the question we have to ask ourselves is what would be the original function to give us a derivative of 9.8? Well, it would have to be 9.8 T. And if G of T is 9.8 T, this implies that V of T would be equal to 9.8 T plus C. Now remember from our problem, we're also told that V of 0 equals -5, OK? So at. V of 0 equals -5 or initial velocity and the time. That means the time would be equal to 0 and the velocity would be equal to -5, which means that we can use that to solve for our constant because no, applying it to our problem V of 0, -5 equals 9.8 multiplied by T 0 plus C. This tells us then that C would have to be equal to -5, and if C equals -5. Then our velocity V of T equals 9.8 T minus 5. Now that we have a function for our velocity, we can use it to help us find a position function. Because now again by the the mean value theorem. OK. It follows that our position functions of T will be equal to H of T plus C, where the derivative of S of T will be equal to the derivative of H of T. That derivative is our velocity. OK, so since the derivative, or rather let's say if. Our derivative is equal to 9.8 T + 5. The question is what can we differentiate to get this result? Well, if that's the case, then H of T would have to be 4.9 T squared minus 5T. And if that is the value of H of T, then by the mean value theorem at the point in our problem we were told our initial initial position equals 2, in other words, at S of 0 equals 2. Which means that T will be equal to 0 and S will be equal to 2. We can use this to solve for our constant and thus for S of T. Because no, using that for a constant, then this implies that here S of T2 is going to be equal to H of T 4.9 multiplied by 0 squared. OK, minus 5 multiplied by 0, thus C. Both of these will be equal to 0, so this implies that C equals 2. And if C equals 2. Then that means SFT. Is going to be equal to age of T 4.9 T square minus 5T plus 2. Thus, this is the function that represents the position of an object with an acceleration A of 9.8, an initial position S of 0 equals 2, and an initial velocity V of 0 equals -5. Thanks a lot for watching everyone. I hope this video helped.

Finding Position from Velocity or Acceleration

Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.

a = 9.8, v(0) = −3, s(0) = 0

Key Concepts

Integration

Initial Conditions

Position Function

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