Classical Mechanics - Acceleration

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻ⁿ where β and n are constants and x is the position of the particle. What is the acceleration of the particle as a function of x?

A. −nβ²x⁻²ⁿ⁻¹
B. −nβ²x⁻ⁿ⁻¹
C. −nβ²x⁻ⁿ
D. −β²x⁻ⁿ⁺¹
E. −β²x⁻²ⁿ⁺¹

(GR9677 #44)

Solution:

a = dv/dt = (dv/dx) (dx/dt) =  (dv/dx) v

Given: v(x) = βx⁻ⁿ
dv/dx =  −nβx⁻ⁿ⁻¹

Thus, a = (−nβx⁻ⁿ⁻¹) βx⁻ⁿ = −nβ²x⁻²ⁿ⁻¹

Answer: A