Cracking The Ap Calculus ab Exam 2018

the derivative  of  the velocity    with    respect to  time.

a(t)    =   2t  −   8

At t = 1,

a(1)    =   2   −   8   =   −6

It does not equal zero, so the particle is changing direction at t = 1.

At t = 7,

a(7)    =   14  −   8   =   6

It does not equal zero, so the particle is changing direction at t = 7.

(c) Find the total distance traveled by the particle from time t = 0 to t = 4.

If we want to find the total distance that a particle travels from time a to time b, we need to
evaluate

This means that, over an interval where the particle’s velocity is negative, we multiply the
integral by −1. So, we need to find where the velocity is negative and where it is positive.

We know that the velocity is zero at t = 1 and at t = 7.

We can find that the velocity is positive when t < 1 and when t > 7, and that the velocity is
negative when 1 < t < 7.

Thus, the distance that the particle travels from t = 0 to t = 4 is

(t^2    −   8t  +   7)  dt  −       (t^2    −   8t  +   7)  dt

Evaluating the integrals, we get