Question

A particle moves along X-axis and its displacement at any time is given by $x\left(t\right)=2t^3-3t^2+4t$ in SI units. The velocity of the particle when its acceleration is zero, is

• A. 2.5 $m{s}^{-1}$
• B. 3.5 $m{s}^{-1}$
• C. 4.5 $m{s}^{-1}$
• D. 8.5 $m{s}^{-1}$

Answer 1

The correct option is A 2.5 $m{s}^{-1}$

Displacement of the particle $x\left(t\right)=2t^3-3t^2+4t$ $m$

Thus velocity of particle $v\left(t\right)=\frac{d}{dt}x\left(t\right)=6t^2-6t+4$ $m/s$

Acceleration of particle $a\left(t\right)=\frac{d}{dt}v\left(t\right)=12t-6$ $m/s^2$

Let acceleration is zero at the instant $t_o$ i.e. $a\left(t_o\right)=0$

$\therefore$ $12t_o-6=0$ $⟹t_o=0.5$ s

Thus velocity at the instant $t_o$, $v\left(t_o\right)=6(0.5{)}^2-6(0.5)+4=2.5$ $m/s$