Q. For a particle moving in a straight line, if time $t$ be regarded as a function of velocity $v$, then the rate of change of the acceleration a is given by
A) $a^{2}\frac{d^{2}t}{dv^{2}}$
B) $a^{3}\frac{d^{2}t}{dv^{2}}$
C) $- a^{3}\frac{d^{2}t}{dv^{2}}$
D) None of these
Explanation
(c)
Let $t = f(v,\ then)$
$\frac{dt}{dv} = f^{'}(v) \Rightarrow \frac{dv}{dt} = \frac{1}{f^{'}(v)}$
$\Rightarrow \ \ a = \frac{1}{f^{'}(v)} \Rightarrow af^{'}(v) = 1\$
$\Rightarrow \ \ af^{''}(v)\frac{dv}{dt} + \frac{da}{dt}f^{'}(v) = 0$
$\Rightarrow \ \ a^{2}f^{''}(v) + \frac{da}{dt} \times \frac{1}{a} = 0 \Rightarrow \frac{da}{dt} = - a^{3}\frac{d^{2}t}{dv^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.