A particle of unit mass undergoes one dimensional motion such that its velocity varies according to

Question

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

Professor PC Thomas Classes · Answer

$a = v \frac{dv}{dx} = (bx^{-2n})(-2nbx^{-2n-1}) = -2nb^2x^{-4n-1}$.