Question

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

(a) –2nb2x –4n–1

(b) –2b2x –2n+1

(c) –2nb2e –4n+1

(d) –2nb2x –2n–1

Answer 1

Correct option (a) –2nb2x –4n–1

Explanation:

According to question,

v(x) = bx-2n

So, dv/dx = - 2nbx-2n - 1

Acceleration of the particle as function of x,

a = vdv/dx = bx-2n {b(-2n)x-2n -1}

= - 2nb2x-4n - 1