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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

Acceleration (aa) is defined as the rate of change of velocity with respect to time (a=dvdta = \frac{dv}{dt}). When velocity is given as a function of position v(x)v(x), we use the chain rule to express acceleration as: a=dvdx⋅dxdt=vdvdxa = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}

  1. Differentiate velocity with respect to position: Given v=βx−2nv = \beta x^{-2n} dvdx=ddx(βx−2n)=β(−2n)x−2n−1=−2nβx−(2n+1)\frac{dv}{dx} = \frac{d}{dx}(\beta x^{-2n}) = \beta(-2n)x^{-2n-1} = -2n\beta x^{-(2n+1)}

  2. Substitute into the acceleration formula: a=v(dvdx)a = v \left( \frac{dv}{dx} \right) a=(βx−2n)⋅(−2nβx−(2n+1))a = (\beta x^{-2n}) \cdot (-2n\beta x^{-(2n+1)})

  3. Simplify: a=−2nβ2x−2n+(−2n−1)a = -2n\beta^2 x^{-2n + (-2n - 1)} a=−2nβ2x−4n−1a = -2n\beta^2 x^{-4n - 1}

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