Question

A body starts from the origin and moves along the $x$-axis such that velocity at any instant is given by $\left(4 t^{3}-2 t\right)$, where $t$ is in second and velocity is in $m / s$. What is the acceleration of the particle, when it is $2 \,m$ from the origin?

• A. $28 \,m/s^2$
• B. $22\, m/s^2$
• C. $12 \,m/s^2$
• D. $10 \,m/s^2$

Answer 1

Answer: (B)

Given that $v=4 t^{3}-2 t$
$x =\int v d t, x=t^{4}-t^{2}+C,$ at $ t=0, x=0 $
$\Rightarrow C =0$
When particle is $2\, m$ away from the origin, then
$2=t^{4}-t^{2} \Rightarrow t^{4}-t^{2}-2=0$
$\Rightarrow \left(t^{2}-2\right)\left(t^{2}+1\right) =0 $
$\Rightarrow t =\sqrt{2} s$
$ a =\frac{d v}{d t}=\frac{d}{d t}\left(4 t^{3}-2 t\right) $
$a =12 t^{2}-2 $
For $ t =\sqrt{2} sec$
$ a =12 \times(\sqrt{2})^{2}-2$
$=22\, m / s ^{2}$