# 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?

Jun 12, 2018

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

#### Explanation:

If the acceleration of the particle is $a \left(t\right) = 3 {t}^{2} - 2 t$, then the velocity of the particle is:

$\int a \left(t\right) \mathrm{dt} = \int \left(3 {t}^{2} - 2 t\right) \mathrm{dt} = {t}^{3} - {t}^{2} + C$

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

${0}^{3} - {0}^{2} + C = 0$

$C = 0$

So our equation simplifies to $v \left(t\right) = {t}^{3} - {t}^{2}$

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

$v \left(t\right) = 0$

${t}^{3} - {t}^{2} = 0$

${t}^{2} \left(t - 1\right) = 0$

$t = 0 \text{ " and " } t = 1$

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):

$v \left(\textcolor{red}{\frac{1}{2}}\right) = {\left(\textcolor{red}{\frac{1}{2}}\right)}^{3} - {\left(\textcolor{red}{\frac{1}{2}}\right)}^{2} = \frac{1}{8} - \frac{1}{4} = - \frac{1}{8}$

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

$v \left(\textcolor{b l u e}{2}\right) = {\left(\textcolor{b l u e}{2}\right)}^{3} - {\left(\textcolor{b l u e}{2}\right)}^{2} = 8 - 4 = 4$

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

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