Science:Math Exam Resources/Courses/MATH100/December 2013/Question 03 (b)/Solution 1

When the particle is moving in the negative direction, its velocity is negative. Therefore we must look for the $t$ for which ${\frac {ds}{dt}}=f'(t)<0$:

${\begin{aligned}s&=t^{3}-{\frac {21}{2}}t^{2}+30t\\{\frac {ds}{dt}}&=3t^{2}-21t+30\end{aligned}}$

Setting the derivative (i.e., the velocity function) equal to zero, we obtain $\displaystyle 0=3t^{2}-21t+30=3(t-2)(t-5)$. Hence the velocity is zero when $\displaystyle t=2,\,5$.

Plugging in values tells us that ${\frac {ds}{dt}}=f'(t)<0$ on the interval ${\color {blue}t\in (2,5)}$, where $t$ is in seconds as stated above.