4.1 , newtons laws of motion — Physics STD 11 Science — Question

from Newton's 2nd law, $F=$ ma

$\Rightarrow m a=\left(x^{3}-3 x\right)$

$\Rightarrow 1 kg \times a =\left( x ^{3}-3 x \right)[ mass$ of body $, m =1 kg ]$

$\Rightarrow a=\left(x^{3}-3 x\right)$

acceleration is the rate of change of velocity with respect to time.

i.e., $a=d v / d t=v d v / d x$

$\Rightarrow vdv / dx =\left( x ^{3}-3 x \right)$

$\underset{\rightarrow}{\int}_{u}^{v} v d v=\int_{1}^{3} x^{3}-3 x d x$

$\Rightarrow\left(v^{2}-u^{2}\right) / 2=\left[\frac{x^{4}}{4}-\frac{3}{2} x^{2}\right]_{1}^{3}$

as body is at rest at $x=1,$ so $u=0$

$\Rightarrow v ^{2} / 2-0=\left[3^{4} / 4-3 / 2(3)^{2}\right]-\left[1^{4} / 4-3 / 2(1)^{2}\right]$

$\Rightarrow v ^{2} / 2=[20.25-13.5]-[0.25-1.5]$

$\Rightarrow v ^{2} / 2=6.75+1.25=8$

$\Rightarrow v ^{2}=16$

$\Rightarrow v=4$

Therefore the velocity of the body at $x=3$ can be $4 m / s .$