Question

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

Answer 1

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

Explanation:

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

`inta(t)dt = int(3t^2-2t)dt = t^3 - t^2 + C`

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

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

`C = 0`

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

Every point where the velocity is `0` is a potential turning point:

`v(t) = 0`

`t^3 - t^2 = 0`

`t^2(t-1) = 0`

`t = 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=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 = -1/8`

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

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

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

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

Final Answer