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 constant and $x$ is the position of the particle. The acceleration of the particle as the function of $x$, is given by:
A) $-2{\beta ^2}{x^{ - 2n + 1}}$
B) $-2n{\beta ^2}{x^{ - 4n + 1}}$
C) $-2n{\beta ^2}{x^{ - 2n - 1}}$
D) $-2n{\beta ^2}{x^{ - 4n - 1}}$

Answer 1

In order to find the value of acceleration in terms of $x$ from the velocity as a function of $x$, we should be able to differentiate velocity with respect to $x$. For that add $\dfrac{{dx}}{{dx}}$ in the equation for the acceleration and do the further differentiation.

Complete step by step solution:
From the question, we know that,
A particle undergoes one-dimensional motion such that the velocity of the object varies according to:
$v(x) = \beta {x^{ - 2n}}$
Where, $\beta$ and $n$ are constant
$x$ is the position of the particle
We are asked to find the acceleration of the particle in terms of $x$
We know, acceleration, $a = \dfrac{{dv}}{{dt}}$
We know that the acceleration is a function of velocity and time.
Hence, we need the acceleration as a function of $x$, we need to multiply the equation with $\dfrac{{dx}}{{dx}}$
That is, $a = \dfrac{{dv}}{{dt}} \times \dfrac{{dx}}{{dx}}$ ……………………. (1)
We know, velocity,$v = \dfrac{{dx}}{{dt}}$
Thus, equation (1) will become,
$a = v\dfrac{{dv}}{{dx}}$……………(2)
Now we are finding the value of $\dfrac{{dv}}{{dx}}$
We are given velocity as function of $x$,
$v(x) = \beta {x^{ - 2n}}$
Differentiating $v$ with respect to $x$, we get,
$\dfrac{{dv}}{{dx}} = - 2n\beta {x^{ - 2n - 1}}$
Now we have the value of both terms $\dfrac{{dv}}{{dx}}$ and $v$ for the equation (2)
Applying these values in the equation (2), we get,
$a = v\dfrac{{dv}}{{dx}}$
$\Rightarrow a = \beta {x^{ - 2n}} \times \left( { - 2n\beta {x^{ - 2n - 1}}} \right)$
$\Rightarrow a = - 2n{\beta ^2}{x^{ - 2n - 1 - 2n}}$
$\Rightarrow a = - 2n{\beta ^2}{x^{ - 4n - 1}}$
That is, the acceleration of the particle which have a velocity, $v(x) = \beta {x^{ - 2n}}$ in terms of $x$ given by,
$a = - 2n{\beta ^2}{x^{ - 4n - 1}}$

So the final answer is option (D), $- 2n{\beta ^2}{x^{ - 4n - 1}}$.

Note: The velocity of an object is the rate of change of that object’s position with respect to a frame of reference, and is a function of time.
In mechanics, acceleration is defined as the rate of change of the velocity of the object with respect to time. Acceleration has the dimensions of velocity $\left( {L{T^{ - 1}}} \right)$ divided by time, i.e. $L{T^{ - 2}}$.