How many values of t does the particle change direction if a particle moves with acceleration a(t)=3t^2-2ta(t)=3t22t and it's initial velocity is 0?

1 Answer
Jun 12, 2018

The particle changes direction for one value of t: t=1t=1.

Explanation:

If the acceleration of the particle is a(t) = 3t^2 - 2ta(t)=3t22t, then the velocity of the particle is:

inta(t)dt = int(3t^2-2t)dt = t^3 - t^2 + Ca(t)dt=(3t22t)dt=t3t2+C

Since the initial velocity is 0 (when t=0):

0^3 - 0^2 + C = 00302+C=0

C = 0C=0

So our equation simplifies to v(t) = t^3 - t^2v(t)=t3t2

Every point where the velocity is 00 is a potential turning point:

v(t) = 0v(t)=0

t^3 - t^2 = 0t3t2=0

t^2(t-1) = 0t2(t1)=0

t = 0 " " and " " t=1t=0 and t=1

To check whether the particle changes directions at each of these points, we need to pick test points to check the intervals between them (we don't have to check before t=0t=0 because that is when the particle starts moving):

v(color(red)(1/2)) = (color(red)(1/2))^3 - (color(red)(1/2))^2 = 1/8 - 1/4 = -1/8v(12)=(12)3(12)2=1814=18

So in the interval from t=0t=0 to t=1t=1, the particle moves in the negative direction.

v(color(blue)2) = (color(blue)2)^3 - (color(blue)2)^2 = 8-4 = 4v(2)=(2)3(2)2=84=4

So in the interval beyond t=1t=1, the particle moves in the positive direction.

Therefore, the particle changes direction at t=1t=1.

Final Answer