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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

-2n\beta ²x⁻²ⁿ⁻¹

B

-2n\beta ²x⁻⁴ⁿ⁻¹

C

-2\beta ²x⁻²ⁿ⁺¹

D

-2n\beta ²x⁻⁴ⁿ⁺¹

Step-by-Step Solution

Given the velocity as a function of position v(x)=βx−2nv(x) = \beta x^{-2n}. The acceleration aa is defined as the rate of change of velocity with respect to time, a=dvdta = \frac{dv}{dt}. Using the chain rule, acceleration can be expressed in terms of position as: a=dvdx⋅dxdt=vdvdxa = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}

  1. Find dvdx\frac{dv}{dx}: dvdx=ddx(βx−2n)\frac{dv}{dx} = \frac{d}{dx}(\beta x^{-2n}) dvdx=β(−2n)x−2n−1=−2nβx−(2n+1)\frac{dv}{dx} = \beta (-2n) x^{-2n - 1} = -2n\beta x^{-(2n+1)}

  2. Calculate Acceleration (aa): Substitute vv and dvdx\frac{dv}{dx} into the acceleration equation: a=(βx−2n)⋅(−2nβx−(2n+1))a = (\beta x^{-2n}) \cdot (-2n\beta x^{-(2n+1)}) a=−2nβ2⋅x−2n+(−2n−1)a = -2n \beta^2 \cdot x^{-2n + (-2n - 1)} a=−2nβ2x−4n−1a = -2n \beta^2 x^{-4n - 1}

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