A particle $P$ starts at the point $O$ and travels in a straight line. At time $t$ seconds after leaving $O$ the velocity of $P$ is v $m/s$, where $v = 0.75t^2 − 0.0625t^3$. Find
(i) the positive value of $t$ for which the acceleration is zero, [3]
(ii) the distance traveled by $P$ before it changes its direction of motion. [5]
I solved (i) and the answer is $t=8$ seconds.
For (ii) I integrated $v = 0.75t^2 − 0.0625t^3$, but stuck after that. Please help.
You are right for (i), now for (ii), you want to know when P changes it direction of motion, i.e. when $v$ is negative. So you just got to solve : $$v(t) < 0 \iff (t>0 \text{ and } \frac{v(t)}{t} < 0) \text{ or } (t<0 \text{ and } \frac{v(t)}{t} > 0) $$
This is factoring by a root of the $v$ polynomial. Here, clearly $0$ is a root so you can factor $v$ by $t-0=t$.
Clearly, $t < 0$ isn't relevant, so you just got to solve : $$0.75t - 0.0625t^2 < 0$$ Which is a $2$nd degree.
More generally, when you got such a problem, try to factor your polynomial, until you can get every single root of it, and then just find the sign of it.