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 = 32, v(0) = 20, s(0) = 5

Accepted Answer

Welcome back, everyone. In this problem, the acceleration of an object moving along a straight line is A equals 32. Its initial position and velocity are S of 0 equals 8 and V of 0 equals 12 respectively. Using the second corollary of the mean value theorem, determine the object's position at a time T. Now how are we going to use the second corollary to help us figure out the object's position at a time T? Well, recall that the corollary tool of the mean value theorem says that if the derivative of our function F X is equal to the derivative of some other function G of X at each point X. In an open interval AB, OK. Then There exists a constant C, OK. There exists a constant C. Such that FFX is equal to GFXX. Plus C for all values of X in our open interval AB, OK? So now for us to figure out what the object's position is, we'll need to make use of this corre. Now if we think about it, we're talking about derivatives, and the derivative of position is the velocity, but we don't have a formula for velocity. However, we know that the acceleration is 32 and the acceleration is the derivative of the velocity. So we could apply the second corollary in that way as well. In other words, if we use it to find the function. A4 or the velocity function for an acceleration of 32, then we should be able to take that velocity function and use it to find our position function given the initial position. So let's first try to find the velocity function and then the position function, OK? Now, let me put that in blue here, finding our velocity. Since A is the derivative of the velocity with respect to time T K D V D T K, then this means that the derivative of a velocity is equal to 32. By the mean value theorem by this corollary, then we'll need to ask ourselves what function when we differentiate it gives us a derivative of 32. Well, if the derivative is 32, that means the velocity must be equal to 32 T. Now in this case. OK, then that would imply that. At the point OK, where we have 0 equals 12, we could use this to slur or constant and thus sulfur or velocity function, OK? Because now at V 0 equals 12, we know that 0, sorry, T is equal to 0. And V is equal to 12, then applying the second color here, we are saying that our function V of T is going to be equal to the derivative 32 T plus C, which now means that 12 is going to be equal to 32 multiplied by 0 plus C. In that case, that means that C is gonna be equal to 12, which means then that our velocity function V of T is going to be equal to 32 T plus 12. OK? So this is our velocity function. Now we can use this as the derivative to figure out our position function, OK? Because now, let me put that in red here. Since our velocity is the derivative of the positions with respect to time, OK. Then that means, OK, that means that the derivative with respect to time is gonna be equal to 32 T + 12. No, we're, we're basically repeating the process again. If this is the derivative of our function, what would we differentiate to get a result of 32 T + 12? Then it would have to follow that the original function H of T. Would be equal, OK, to 16 T2 plus 12 T. 16, the derivative of 16 T2 is 32 T and the derivative of 12 is 12 T. Now, again, That means since this is the derivative. At the point where S of 0. OK. SO 0 equals 8. Which means that T equals 0 and S equals 8. If we take our functions of T equals V of T + C, H of T + C. And substitute what we know, then that means S, which is 8, is going to be equal to 16 T squared plus 12T, OK, plus. See Which, as we said, T is 0, so I should really write this as 16 multiplied by 0 squared plus 12 multiplied by 0 + 8. I see, sorry. If that's the case, both of these will be equal to 0, so that means C equals 8. And now, with C being equal to 8, this would imply then that S of T is equal to 16 T 2 + 12 T + 8. In other words, the function that represents the position of the object with an acceleration A of 32, and initial positions of 0 equals 8, and initial velocity V of 0 equals 12 is S of T equals 16 T2 + 12 T + 8. Thanks 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 = 32, v(0) = 20, s(0) = 5

Key Concepts

Integration

Initial Conditions

Position Function

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