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. -2nβ²e^-4n+1
• B. -2nβ²x^-2n-1
• C. -2nβ²x^-4n-1
• D. -2nβ²x^-2n+1

Answer 1

The correct option is C

-2nβ²x^-4n-1

$v\left(x\right)=\beta x^{-2n}a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}=\frac{dv}{dx}.v\frac{dv}{dx}=-2n\beta x^{-2n-1}\therefore a=-2n\beta x^{-2n-1}.\beta x^{-2n}=2n{\beta }^2{x}^{-4n-1}$