Question

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

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

Answer 1

Answer: (B)

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