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

The Correct Option is D

To find the acceleration of the particle as a function of its position \(x\), we start with the given velocity function:

v(x) = \beta x^{-2n}

where \(\beta\) and \(n\) are constants.

In kinematics, acceleration \(a\) is defined as the derivative of velocity \(v\) with respect to time \(t\). However, since velocity is given as a function of position, we can use the chain rule to find acceleration in terms of position:

a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}

First, calculate the derivative of \(v(x)\) with respect to \(x\):

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

Since v(x) = \beta x^{-2n}, we substitute \(v\) and \(\frac{dv}{dx}\) into the expression for acceleration:

a = \left(\beta x^{-2n}\right) \left(-2n\beta x^{-2n-1}\right)

This simplifies to:

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

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

Therefore, the acceleration of the particle as a function of \(x\) is:

-2n\beta^2 x^{-4n-1}

This matches the given correct answer.