Question

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

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

Answer 1

The correct option is C $22m/s^2$

Find the displacement equation of the body.
Given velocity of body
$v=4t^3-2t$
Therefore acceleration
$a=\frac{dv}{dt}=12t^2-2\dots \left(i\right)$
and
$x={\int }_0^{t}vdt={\int }_0^t(4t^3-2t)dt$
$x=t^4-t^2\dots \left(ii\right)$
Find the acceleration of the body.
When particle is at $2m$ from the origin, then, $x=2$, from equation $\left(ii\right)$
$t^4-t^2=2$
$\Rightarrow t^4-t^2-2=0$
$\Rightarrow (t^2-2)(t^2+1)=0$
$\Rightarrow t=\sqrt{2}\sec$
Acceleration at $t=\sqrt{2}\sec$ given by
$a=12t^2-2=12\times 2-2$
$a=22m/s^2$ (From equation $\left(i\right)$)
Final answer: $\left(b\right)$