### Neet Solved Paper 2015 Question 3

##### 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

#### Options:

A) $ -2n{{\beta }^{2}},{x^{-2n-1}} $

B) $ -2n{{\beta }^{2}},{x^{-4n-1}} $

C) $ -2{{\beta }^{2}}{x^{-2n+1}} $

D) $ -2n{{\beta }^{2}},{e^{-4n+1}} $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

Given, $ v=\beta {x^{-2n}} $ $ a=\frac{dv}{dt}=\frac{dx}{dt}.\frac{dv}{dx} $

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

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