Question

A body starts from the origin and moves along the axis such that the velocity at any instant is given by $v=4t^3-2t$ where t is in second and the velocity is in $m/s$ . Find the acceleration of the particle when it is at a distance of $2m$ from the origin.

• A. $28m{s}^{-2}$
• B. $22m{s}^{-2}$
• C. $12m{s}^{-2}$
• D. $10m{s}^{-2}$

Answer 1

Answer: (B)

Velocity $v=4t^3-2t$ ... (i)
$\frac{dx}{dt}=4t^3-2t$
On integration, we get
$x=2=t^4-t^2={\alpha }^2-\alpha$ ... (ii)
( $if{t}^2=\alpha$)
${\alpha }^2-\alpha -2=0$
$(\alpha -2)(\alpha +1)=0$
$\alpha =2$
$\alpha =-1$ , which is not possible
$t^2=\alpha =2$ or $t=\sqrt{2}$
Differentiating Eq. (i), w.r.t. $t$
$\frac{dv}{dt}=12t^2-2$
$\alpha =12\times 2-2=22m{s}^{-2}$