Question

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v\left(x\right)=$ $b{x}^{-2n}$, where $b$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$, is given by

Answer 1

Given:
$v\left(x\right)=b{x}^{-2n}$

Differentiating both sides:
$\frac{dv}{dx}=$ $-2nb{x}^{-2n-1}$

We know that, acceleration of the particle as a function of $x$ is given as:
$a\left(x\right)=v\frac{dv}{dx}=\left(b{x}^{-2n}\right)[-2bn{x}^{-2n-1}]$

$\Rightarrow$ $a\left(x\right)=$ $-2n{b}^2{x}^{-4n-1}$

Hence, the option $\left(A\right)$ is correct.