Question

A particle is moving in a straight line such that its velocity varies according to $v\left( x \right) = 2{x^{ - 3m}}$, where $m$ is a constant and $x$ is the position of the particle. The acceleration of the particle as a function of $x$ is given by-
(1) $ - 12m{x^{ - 6m + 1}}$
(2) $ - 12m{x^{ - 3m - 1}}$
(3) $ - 8m{x^{ - 3m - 1}}$
(4) $ - 12m{x^{ - 6m - 1}}$

Answer 1

Hint: We know that the acceleration is described as the rate of variation of the velocity of an object. If the position of the object is $x$ then the velocity of the object in the derivative form is given by,
$v\left( x \right) = \dfrac{{dx}}{{dt}}$
And the acceleration is the derivative of the velocity w.r.t. the time. So, the acceleration of the object in the derivative form is given by,
Acceleration $a\left( x \right) = \dfrac{{dv\left( x \right)}}{{dt}}$
Or, $\begin{array}{l}
a\left( x \right) = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right)\\
a\left( x \right) = \dfrac{{{d^2}x}}{{d{t^2}}}
\end{array}$

Complete step by step answer:
The expression of variation of the velocity of a particle $v\left( x \right) = 2{x^{ - 3m}}$
Where $m$ is a constant and $x$ is the position of the particle.
In order to find the acceleration of this particle, we have to find the derivative of the function$v\left( x \right) = 2{x^{ - 3m}}$.
So, taking derivatives of both sides of the expression w.r.t. time $t$ we get,
\[\dfrac{d}{{dt}}\left\{ {v\left( x \right)} \right\} = \dfrac{d}{{dt}}\left( {2{x^{ - 3m}}} \right)\]
We know that acceleration is $a\left( x \right) = \dfrac{{dv\left( x \right)}}{{dt}}$.
So substituting this in the expression we have,
$a\left( x \right) = 2\dfrac{d}{{dt}}\left( {{x^{ - 3m}}} \right)$
We can write $\dfrac{d}{{dt}}\left( {{x^{ - 3m}}} \right)$ as-
$\begin{array}{l}
\dfrac{d}{{dt}}\left( {{x^{ - 3m}}} \right) = \left( {\dfrac{d}{{dx}}\left( {{x^{ - 3m}}} \right)} \right) \times \dfrac{{dx}}{{dt}}\\
 = \left( {\dfrac{d}{{dx}}\left( {{x^{ - 3m}}} \right)} \right) \times v\left( x \right)
\end{array}$
Substituting this value into the expression we get,
$a\left( x \right) = 2\left( {\dfrac{d}{{dx}}\left( {{x^{ - 3m}}} \right)} \right) \times v\left( x \right)$
Substitute $v\left( x \right) = 2{x^{ - 3m}}$ in the expression, we get,
$a\left( x \right) = 2\left( {\dfrac{d}{{dx}}\left( {{x^{ - 3m}}} \right)} \right) \times 2{x^{ - 3m}}$
And we know that $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ so,
$a\left( x \right) = 2 \times \left( { - 3{\rm{m}} \times {{\rm{x}}^{ - 3m - 1}}} \right) \times 2{x^{ - 3m}}$
Solving this we get,
$\begin{array}{l}
a\left( x \right) = - 12m{x^{ - 3m - 1 - 3m}}\\
a\left( x \right) = - 12m{x^{ - 6m - 1}}
\end{array}$
Therefore, the acceleration of the particle as a function of $x$ is $a\left( x \right) = - 12m{x^{ - 6m - 1}}$ and the correct option is (4).

Note: The acceleration of the particle is calculated w.r.t. time but the expression of the velocity is given in form of position $x\left( t \right)$ so we have to rearrange it in following way-
$\dfrac{{dv\left( x \right)}}{{dt}} = \dfrac{{dv\left( x \right)}}{{dx}} \times \dfrac{{dx}}{{dt}}$