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)

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

Answer 2

According to question, velocity of unit mass varies as
$v(x)=\beta x^{-2n}$ $50mm$ $\quad$ $\ldots\left(i\right)$
$\frac{dv}{dx}$ $=-2n\beta x^{-2n-1}$ $40\,mm$ $\quad$ $\ldots\left(ii\right)$
Acceleration of the particle is given by
$a=\frac{dv}{dt}=\frac{dv}{dx} \times \frac{dx}{dt}=\frac{dv}{dx} \times v$
Using equation (i) and (ii), we get
$a =(-2n\beta x^{-2n-1}) \times (\beta x^{-2n})$
$= - 2n\beta^2 x^{-4n-1}$