Velocity Integral Question

Question

Suppose that a particle travels with velocity v(t) = p81 − (t − 9)2

meters per second between t = 0 and t = 18 seconds.

1. Write a function for how far the particle has traveled after x seconds. Call

this function P(x).

2. How far has the particle traveled after 9 seconds.

3. Suppose that there is another particle with position P(x/2). At what

velocity is that particle traveling at time x = 5?

Answer 1

v(t) = sqrt(81 - (t - 9)^2) graphs like this:

Because it never goes negative we don't need to worry about the particle backing up and, in so doing, traveling some additional distance. The area under the curve is the position and the area under the curve is the integral of velocity:

At x = 9, the area is 1/4 the area of a circle (πr²) which, since r = 9, would be 81π/4

Assuming the second particle had the same position function then, since velocity is the derivative of position the velocity would be:

v(x/2) = P'(x/2) = 1/2(√(81 - (x/2 - 2))²) and you can plug in x = 5