Question

A particle moves along a straight line such that its displacement(s) at any time(t) is given by $S=t^3-3t^2+2$. The velocity of the particle when its acceleration zero is?

• A. $-2$ units
• B. $-8$ units
• C. $0$ units
• D. $-9$ units

Answer 1

The correct option is C. $0$ units

Displacement is given as $=(t^3-3t^2+2)⟹\left(1\right)$

We know that $a=\frac{d^{2}s}{d{t}^2}$

Differentiating (1 ) once with respect to t we get

$v=\frac{ds}{dt}=\frac{d}{dt}(t^3-3t^2+2)=3t^2-6t$

Differentiating once again we get

$\frac{d^{2}s}{d{t}^2}=\frac{d}{dt}(3t^2-6t)=6t-6$

Given condition is $\frac{d^{2}s}{d{t}^2}=6t-6=0$

$t=1s$

Therefore,

$v\left(1\right)=\left(3\right(1{)}^2-6\times 1)=-3m/s$.