32. 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 the function of x, is given by:

32. 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 the function of x, is given by:

(a)  \( -2nb^{2}x^{-4n-1}  \)

(b)  \( -2b^{2}x^{-2n+1}  \)

(c)  \( -2nb^{2}x^{-4n+1}  \)

(d)  \( -2nb^{2}x^{-2n-1}  \)

Answer :  (a)  \( -2nb^{2}x^{-4n-1}  \)

\( v = b x^{-2n} \)

\( a = \)\(\frac{dv}{dt}\)\( = b(-2n)x^{-2n-1} \)\(\frac{dx}{dt} \)

\( = b(-2n)x^{-2n-1} v \)

\( = b(-2n)x^{-2n-1} \, b(x^{-2n}) \)

\( = -2n b^2 x^{-4n-1} \)