Question

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to
$v\left(x\right)=\beta x^{-2n}$
where $\beta$ 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

• A. $-2n{\beta }^2{x}^{-4n-1}$
• B. $-2{\beta }^2{x}^{-2n+1}$
• C. $-2n{\beta }^2{x}^{-4n+1}$
• D. $-2n{\beta }^2{x}^{-2n-1}$

Answer 1

The correct option is A $-2n{\beta }^2{x}^{-4n-1}$

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

As we know that

$a=v\frac{dv}{dx}$

$\therefore a=v\frac{dv}{dx}=v(-2\beta n{x}^{-2n-1})$

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

$\Rightarrow a=-2n{\beta }^2{x}^{-4n-1}$

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