Physics - Kinematics Question with Solution | TestHub

Acceleration can be written as, $a=v\frac{dv}{dx}$.

Given: $a=-2x$(-ve sign is due to retardation)

Therefore,

$$\begin{gathered} v\frac{dv}{dx}=-2x \\ \Rightarrow vdv=\left(-2x\right)dx \end{gathered}$$

Integrating both sides, we get

$$\begin{gathered} \frac{1}{2}\left(v_f^2-v_i^2\right)=-2\left(\frac{x^2}{2}\right)=x^2 \\ \Rightarrow \left|\frac{1}{2}m\left(v_f^2-v_i^2\right)\right|=m{x}^2 \end{gathered}$$

$\Rightarrow \left|\Delta KE\right|=0.01x^2={\left(\frac{10}{x}\right)}^{-2}$

Hence, $n=2$.