Question

A body starts from origin and moves along x-axis such that at any instant velocity is $v_t=4t^3-2t$ where $t$ is in second and $v_t$ is in ${ms}^{-1}$. The acceleration of the particle when it is $2m$ from the origin is

• A. $28{ms}^{-2}$
• B. $22{ms}^{-2}$
• C. $12{ms}^{-2}$
• D. $10{ms}^{-2}$

Answer 1

The correct option is B $22{ms}^{-2}$

Given,

$v_t=4t^3-2t$

Acceleration, $a=\frac{d{v}_t}{dt}$

$a=12t^2-2$. . . . . . . . . . .(1)

displacement, $ds=v_{t}dt$

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

$t^4-t^2=2$

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

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

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

$t^2=2$

$t=\pm \sqrt{2}$

So, $t=\sqrt{2}$

From equation (1),

$a=12(\sqrt{2}{)}^2-2$

$a=24-2$

$a=22m/s^2$

The correct option is B.