ap calculus frq

Question

A particle moves along a straight line. For 0 ≤ t ≤ 5, the position of the particle is given by

s(t)=3t+4t^2-t^3.

 a) What is the average velocity of the particle over the interval 0 ≤ t ≤ 4?

 b) Find a general expression for the velocity of the particle as a function of time, v(t), and determine the value of v at t = 2. Is the particle traveling to the left or the right at this point? Justify your answer. 

c) Find all times t during the interval 0 ≤ t ≤ 5 that the particle changes direction. Justify your answer.

 d) At t = 4, is the speed of the particle increasing or decreasing? Justify your answer.

Answer 1

a) Average velocity over [0,4] = [s(4) - s(0)] / (4 - 0) = 12/4 = 3

b) s'(t) = v(t) = 3 + 8t - 3t^2

v(2) = 3 + 16 - 12 = 7

The particle is traveling to the right because the velocity is positive.

c) 3 + 8t - 3t^2 = 0

Using quadratic formula, t = 3, and t = -1/3 (these are the critical numbers, and it's where the particle could POSSIBLY switch directions.

1st derivative test:

s ' (-1) = negative

s ' (0) = positive

s ' (4) = negative

Over the interval [0,5], the particle changes directions only at t = 3 seconds, which is when the velocity switches from positive to negative. It also switches directions at t = -1/3 seconds, but that is not in the interval.

d) s''(t) = v'(t) = a(t) = 8 - 6t

a(4) = negative

Because the acceleration at t = 4 is a negative number, the velocity is decreasing. However, the velocity is also negative at t = 4. Therefore, if the velocity is negative and decreasing, that means that the speed is actually increasing because speed is the absolute value of the velocity.