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Question: A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to ...

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) =βx−2n, = \beta {x^{ - 2n}},where β\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. −2β2x−2n+1 - 2{\beta ^2}{x^{ - 2n + 1}}
B. −2nβ2e−4n+1 - 2n{\beta ^2}{e^{ - 4n + 1}}
C. −2nβ2x−2n−1 - 2n{\beta ^2}{x^{ - 2n - 1}}
D. −2nβ2x−4n−1 - 2n{\beta ^2}{x^{ - 4n - 1}}

Explanation

Solution

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

Complete Step by Step Answer:
Given the mass of the particle m=1  unitm = 1\,\,unit
Velocity v(x)=βx−2nv\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=dvdta = \dfrac{{dv}}{{dt}}
Or, a=dvdt×dxdxa = \dfrac{{dv}}{{dt}} \times \dfrac{{dx}}{{dx}}
Or a=dxdt×dvdxa = \dfrac{{dx}}{{dt}} \times \dfrac{{dv}}{{dx}}
Or, a=vdvdx  [∴v=dxdt].......(i)a = v\dfrac{{dv}}{{dx}}\,\,\left[ {\therefore v = \dfrac{{dx}}{{dt}}} \right].......\left( i \right)
Now, v=βx−2n......(ii)v = \beta {x^{ - 2n}}......\left( {ii} \right)
Differentiating the velocity v with respect to position x.
⇒dvdx=ddx[βx−2n]\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left[ {\beta {x^{ - 2n}}} \right]
=βddx[x−2n]     [∵ β  is  constant]= \beta \dfrac{d}{{dx}}\left[ {{x^{ - 2n}}} \right]\,\,\,\,\,\left[ {\because \,\beta \,\,is\,\,constant} \right]
=β[−2n  x−2n−1]= \beta \left[ { - 2n\,\,{x^{ - 2n - 1}}} \right]
⇒dvdx=−2nβ  x−2n−1.........(iii)\Rightarrow \dfrac{{dv}}{{dx}} = - 2n\beta \,\,{x^{ - 2n - 1}}.........\left( {iii} \right)
From equation (i) and (iii)
a=v  dvdxa = v\,\,\dfrac{{dv}}{{dx}}
a=v(−2nβ  x−2n−1)a = v\left( { - 2n\beta \,\,{x^{ - 2n - 1}}} \right)
a=2nβv  x−2n−1.........(iv)a = 2n\beta v\,\,{x^{ - 2n - 1}}.........\left( {iv} \right)
From equation (ii) and (iv)
a=−2nβ(βx−2n)x−2n−1a = - 2n\beta \left( {\beta {x^{ - 2n}}} \right){x^{ - 2n - 1}}
a=−2nβ2x−2n−2n−1a = - 2n{\beta ^2}{x^{ - 2n - 2n - 1}}
a=−2nβ2x−4n−1a = - 2n{\beta ^2}{x^{ - 4n - 1}}
Hence, option (D) is correct.

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