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 2 m 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

$22m/s^2$

Given v = $4t^3-2t$ .

$\therefore a=\frac{dv}{dt}=12t^2-2$

and $x=\int vdt=\int (4t^3-2t)dt=t^4-t^2$

When particle is at 2m from the origin, we have

$t^4-t^2=2$

$\Rightarrow t^4-t^2-2=0$

$(t^2-2)(t^2+1)=0\Rightarrow t=\sqrt{2}$sec (only possible physical root)

Acceleration at $t=\sqrt{2}$ sec given by,

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