ANSWERS AND EXPLANATIONS TO SECTION II
- A particle moves along the x-axis so that its acceleration at any time t > 0 is given by a(t) = 12t
− 18. At time t = 1, the velocity of the particle is v(1) = 0 and the position is x(1) = 9.
(a) Write an expression for the velocity of the particle v(t).Step 1: We know that the derivative of velocity with respect to time is acceleration, so the
integral of acceleration with respect to time is velocity.∫^ a(t) dt = v(t)
∫^12 t − 18 dt = 6t
(^2) − 18t + C = v(t)
If we plug in the information that at time t = 1, v(1) = 0, we can solve for C.
6(1)^2 − 18(1) + C = 0
−12 + C = 0
C = 12
This means that the velocity of the particle is 6t^2 − 18t + 12.
(b) At what values of t does the particle change direction?
When a particle is in motion, it changes direction at the time when its velocity is zero. (As long
as acceleration is not also zero.) So all we have to do is set velocity equal to zero and solve
for t.
6 t^2 − 18t + 12 = 0
t^2 − 3t + 2 = 0
(t − 2)(t − 1) = 0
t = 1, 2
(c) Write an expression for the position x(t) of the particle.
We know that the derivative of position with respect to time is velocity, so the integral of
velocity with respect to time is position.
∫^ v(t) dt = x(t)
∫ (6t
(^2) − 18t + 12) dt = 2t (^3) − 9t (^2) + 12t + C = x(t)