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^{-2 n} $
$\frac{d v}{d x}=-2 n \beta x^{-2 n-1}$
Acceleration of the particle is given by
$a=\frac{d v}{d t}=\frac{d v}{d x} \times \frac{d x}{d t}=\frac{d v}{d x} \times v$
Using equation (i) and (ii), we get
$a =\left(-2 n \beta x^{-2 n-1}\right) \times\left(\beta x^{-2 n}\right) $
$=-2 n \beta^{2} x^{-4 n-1}$