A particle of unit mass undergoes one-dimensional motion

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A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = bx^–2n where b and n are constants and x is the position of the particle. The acceleration of the particle as d function of x, is given by:

1. –2nb²x^–4n–1
2. –2b²x^–2n+1
3. –2nb²e^–4n+1
4. –2nb²x^–2n–1

Correct Option: A

According to question,
V (x) = bx^–2n
So, dv/dx = –2 nb x^{–2n –1}
Acceleration of the particle as function of x,
a = v (dv/dx) = bx^–2n {b(-2n)X^{2n-1}
= – 2nb²x^–4n–1}