A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻²ⁿ, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is:
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Question

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻²ⁿ, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is:

Options

G1 Physics · Accepted Answer

Step 1: a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx). Step 2: dv/dx = −2nβx⁻²ⁿ⁻¹. Step 3: Multiply by v = βx⁻²ⁿ to get a = −2nβ²x⁻⁴ⁿ⁻¹. Final answer: B.