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 second and velocity is in m/s. What is acceleration of the particle, when it is at distance 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 $22m/s^2$

Velocity, $V=4t^3-2t$ $,$ $t$ is in sec.

Displacement $\int dx=\int vdt$

$\Rightarrow x=\int (t^3-2t)dt$

$=t^n-t^2$

for $x=2m,$ $t=?$

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

$\Rightarrow t^n-2t^2+t^2-2=0$

$\Rightarrow t^2(t^2-2)+1(t^2-2)=0$

$\Rightarrow t^2=2$

$\Rightarrow t=\sqrt{2}sec$

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

$=12\left(2\right)-2$ $(t-\sqrt{2})$

$\Rightarrow a=22m/s^2$

Hence, the answer is $22m/s^2.$