Question

A particle moves along a straight line such that its displacement at any time $t$ is given by $s=(t^3-6t^2+3t+4)\text{m}$. The velocity when the acceleration is zero is

• A. $3\text{m/s}$
• B. $-12\text{m/s}$
• C. $42\text{m/s}$
• D. $-9\text{m/s}$

Answer 1

The correct option is D $-9\text{m/s}$

Given,
$s=t^3-6t^2+3t+4$
$\Rightarrow \frac{ds}{dt}=3t^2-12t+3$
$\Rightarrow v=3t^2-12t+3$.....$\left(1\right)$ $[v=\frac{ds}{dt}]$
$\Rightarrow \frac{dv}{dt}=6t-12$
$\Rightarrow a=6t-12$ $[a=\frac{dv}{dt}]$
$\Rightarrow 6t-12=0$ $[a=0,\text{given}]$
$\Rightarrow t=2\text{s}$ .....$\left(2\right)$
On putting value of $t$ from $\left(2\right)$ in $\left(1\right)$,
$v=3(2{)}^2-12\times 2+3$
$v=-9\text{m/s}$
The velocity of particle when its acceleration is zero is $-9\text{m/s}$.