Question

A particle of unit mass undergoes one dimensional motion such that its velocity varies according to $v(x)=\beta x^{-2n},$ 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. $-2{\beta }^2{x}^{-2n+1}$
• B. $-2n{\beta }^2{e}^{-4n+1}$
• C. $-2n{\beta }^2{x}^{-2n-1}$
• D. $-2n{\beta }^2{x}^{-4n-1}$

Answer 1

Answer: (D)

According to question, velocity of unit mass varies as
$v(x)=\beta x^{-2n}$
$\frac{dv}{dx}=-2n\beta x^{-2n-1}$
Acceleration of the particle is given by
$a=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=\frac{dv}{dx}\times v$
Using equation (i) and (ii), we get
$a=\left(-2n\beta x^{-2n-1}\right)\times \left(\beta x^{-2n}\right)$
$=-2n{\beta }^2{x}^{-4n-1}$