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

Hint:The velocity is defined as rate change of position with respect to time i.e. $v = \dfrac{{dx}}{{dt}}$.
The acceleration of a particle is a function of position and is given by, $a = v\left( {\dfrac{{dv}}{{dx}}} \right)$.

Complete Step by Step Answer:
Given the mass of the particle $m = 1\,\,unit$
Velocity $v\left( x \right) = \beta {x^{ - 2n}}$
The acceleration is defined as the rate change of velocity of a partied with respect to time i.e.
$a = \dfrac{{dv}}{{dt}}$
Or, $a = \dfrac{{dv}}{{dt}} \times \dfrac{{dx}}{{dx}}$
Or $a = \dfrac{{dx}}{{dt}} \times \dfrac{{dv}}{{dx}}$
Or, \[a = v\dfrac{{dv}}{{dx}}\,\,\left[ {\therefore v = \dfrac{{dx}}{{dt}}} \right].......\left( i \right)\]
Now, $v = \beta {x^{ - 2n}}......\left( {ii} \right)$
Differentiating the velocity v with respect to position x.
$ \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left[ {\beta {x^{ - 2n}}} \right]$
$ = \beta \dfrac{d}{{dx}}\left[ {{x^{ - 2n}}} \right]\,\,\,\,\,\left[ {\because \,\beta \,\,is\,\,constant} \right]$
\[ = \beta \left[ { - 2n\,\,{x^{ - 2n - 1}}} \right]\]
$ \Rightarrow \dfrac{{dv}}{{dx}} = - 2n\beta \,\,{x^{ - 2n - 1}}.........\left( {iii} \right)$
From equation (i) and (iii)
$a = v\,\,\dfrac{{dv}}{{dx}}$
$a = v\left( { - 2n\beta \,\,{x^{ - 2n - 1}}} \right)$
$a = 2n\beta v\,\,{x^{ - 2n - 1}}.........\left( {iv} \right)$
From equation (ii) and (iv)
$a = - 2n\beta \left( {\beta {x^{ - 2n}}} \right){x^{ - 2n - 1}}$
$a = - 2n{\beta ^2}{x^{ - 2n - 2n - 1}}$
$a = - 2n{\beta ^2}{x^{ - 4n - 1}}$
Hence, option (D) is correct.

Note: The differentiation of function $y = {x^n}$ is given as, $\dfrac{{dy}}{{dx}} = n\,\,{x^{n - 1}}.$ And also make sure that the concept of one dimensional motion is cleared in your mind.