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A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = bx²n where b and n are constants and x is the position of the particle. The acceleration of the particle as the function of x is given by:

1. -2nb²x⁴n−1
2. -2b²x²n−1
3. -2nb²x⁴n+1
4. -2nb²x²n−1

Correct answer: -2nb²x⁴n−1

Solution

The velocity is given as v(x) = bx²n. Acceleration is obtained using the chain rule: a = v(dv/dx). Differentiating v(x) with respect to x gives dv/dx = 2nbx²n−1. Substituting v and dv/dx into a = v(dv/dx), we get a = bx²n(2nbx²n−1) = -2nb²x⁴n−1.

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