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

Answer 1

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

$v=4t^3-2t\left(given\right)\therefore a=\frac{dv}{dt}=12t^2-2$

and $x=\underset{0}{\overset{t}{\int }}vdt=\underset{0}{\overset{t}{\int }}(4t^3-2t)dt=t^4-t^2$

When particles is at 2m from the origin $t^4-t^2=2$

$\Rightarrow t^4-t^2-2=0(t^2-2)(t^2+1)=0\Rightarrow t=\sqrt{2}sec$

Acceleleration at $t=\sqrt{2}sec$ gievn by,

$a=12t^2-2=12\times 2-2=22m/s^2$