Question

If a particle moves according to the law $s=t^3/3-3t^2+8t$, then the distance covered before it first comes to rest is

• A. 16/3
• B. 8
• C. 10/3
• D. 20/3

Answer 1

Answer: (D)

$\upsilon =\frac{ds}{dt}=\frac{3t^2}{3}-6t+8$
$=t^2-6t+8=\left(t-4\right)\left(t-2\right)$
$a=\frac{d^{2}s}{d{t}^2}=2t-6$
$\upsilon =0$
$\Rightarrow \frac{ds}{dt}=0$
$\Rightarrow \left(t{-}^{}4\right)\left(t-2\right)=0$
$\Rightarrow t=2$ or $4$
Required distance $=s$ when $t=2$
$=\frac{8}{3}-3\left(4\right)+8\left(2\right)$
$=\frac{8}{3}+4=\frac{20}{3}$