Q. The relation between time and distance is $\ t = \alpha x^{2} + \beta x$, where $\alpha$ and $\beta$ are constants. The retardation is
A) $2\alpha v^{3}$
B) $2\beta v^{3}$
C) $2\alpha\beta v^{3}$
D) $2\beta^{2}v^{3}$
Explanation
(a) $\frac{dt}{dx} = 2\alpha x + \beta \Rightarrow v = \frac{1}{2\alpha x + \beta}$
$\because a = \frac{dv}{dt} = \frac{dv}{dx}.\frac{dx}{dt}$
$a = v\frac{dv}{dx} = \frac{- v.2\alpha}{(2\alpha x + \beta)^{2}} = - 2\alpha.v.v^{2} = - 2\alpha v^{3}$
$\therefore$ Retardation $= 2\alpha v^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.