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. Find The acceleration of the particle as a function of $x$, if $n=2$

• A. $-2n{\beta }^2{x}^{-9}$
• B. $-2{\beta }^2{x}^{-3}$
• C. $-2n{\beta }^2{x}^{-7}$
• D. $-2n{\beta }^2{x}^{-5}$

Answer 1

The correct option is A $-2n{\beta }^2{x}^{-9}$

Given data for particle,

Mass, $m=1\text{unit}$
Velocity, $v=\beta x^{-2n}$

As the relation between velocity $v$ and acceleration $a$ for a paricle along the $\text{x-}$axis is given by,
$$a=v\frac{dv}{dx}...\left(1\right)$$ Now,

$\frac{dv}{dx}=\frac{d\left(\beta x^{-2n}\right)}{dx}$

$\Rightarrow \frac{dv}{dx}=-2\beta n{x}^{-2n-1}$

So,

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

$\Rightarrow v\frac{dv}{dx}=(-2\beta n{x}^{-2n-1})\left(\beta x^{-2n}\right)$

$\Rightarrow v\frac{dv}{dx}=-2{\beta }^{2}n{x}^{-4n-1}$

Thus, from equation $\left(1\right)$,

$a=v\frac{dv}{dx}=-2{\beta }^{2}n{x}^{-4n-1}$

Substituting $n=2$ we get,

$a=-2{\beta }^{2}n{x}^{-9}$

Hence, option (a) is the correct answer.