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 = 9.8t + 5, s(0) = 10

Accepted Answer

Welcome back, everyone. In this problem, the velocity of an object moving along a straight line is defined by V equals 9.8 T + 2. Its initial position is S 0 equals 5. Using the second corollary of the mean value theorem, determine the object's position at time T. Now, how is the second corollary of the mean value theorem going to help us find the object's position at a given time T? What does that corollary say? We we recall, OK, it basically tells us that if the derivative of our function F of X is equal to the derivative of some other function G of X at each point X, OK. In an open interval AB. Then There exists a constant C, OK, there exists this constant value. Such that Such that F of X is equal to G of X plus C, OK, for all values of X in the open interval. OK. That is F minus G is a constant function and A B. So if we know the derivative of the function we're looking for S of T, then we should be able to use that to help us solve for a constant C and thus for our function F of X. Now notice here. That for our functions of T that we're looking for, the velocity V is the derivative of the positions with respect to time T. In other words, V equals DSDT. And in this case, since we have an expression for V, it equals 9.8 T + 2, then In our function, if we can let, OK. If we can let S of T equal to some other function, uh, let's call it G of T, OK, plus C based on our second color theorem. So we're just rewriting some of the, the, the terms here, OK. Then we can solve for G of T following from the mean value theorem and thus solve for a C and then S of T. Now if we think about it here. The derivative of G of T is also going to be equal to the velocity, which is 9.8 T, OK, plus 2. So now we have to ask ourselves if this is the derivative of our function GFT, what would our function be? Well, this would imply that our function GFT. Is going to be equal to 4.9 T squared. Plus 2 T. Because the derivative of 4.9 T squared is 9.8T and the derivative of 2 T equals 2. Since that's the case, this would imply then that our position position functions of T equals 4.9 T2 + 2 T + C. Now how are we going to solve for a C? Well, remember, if we go back to our problem statement, we're told that our initial position of our objects of 0 equals 5. So in this case, At the point, OK, where the time equals 0 and S of T is equal to 5, we can substitute it into our equation to solve for C. Because no, this means then. That S of 0, OK, well, I I can say S of 0 equals 5 here. In that case, we're going to get 05. OK. Being equal to 4.9 multiplied by 0 squared, plus 2 multiplied by 0 plus C. Both of these will be equal to 0, so this means that C is equal to 5. And if C is equal to 5, then that means S of T. Will be equal to 4.9 T squared plus 2 T + 5. This is the function that represents the position of the object with a velocity of 9.8 T + 2 and an initial position of 5. Thanks for watching everyone. I hope this video helps.

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 = 9.8t + 5, s(0) = 10

Key Concepts

Integration

Initial Conditions

Definite and Indefinite Integrals

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