Looking for classes? Ksquare Career Institute, Bengaluru →
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:
Explanation
According to question, $V(x) = bx^{-2n}$ So, $dv/dx = -2nbx^{-2n-1}$ Acceleration of the particle as function of x, $a = v dv/dx = bx^{-2n}\{b(-2n)x^{-2n-1}\} = -2nb^{2}x^{-4n-1}$