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. 28 m/s²
• B. 22 m/s²
• C. 12 m/s²
• D. 10 m/s²

Answer 1

The correct option is B 22 m/s²

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

and$x={\int }_0^{t}Vdt={\int }_0^t(4t^3-2t)dt=t^4-t^2$

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

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

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

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