Q. A train accelerates from rest at a constant rate $\alpha$ for distance $x_{1}$ and time $t_{1}$. After that it retards to rest at constant rate $\beta$ for distance $x_{2}$ and time $t_{2}$. Then it is found that
A) $\frac{x_{1}}{x_{2}} = \frac{\alpha}{\beta} = \frac{t_{1}}{t_{2}}$
B) $\frac{x_{1}}{x_{2}} = \frac{\beta}{\alpha} = \frac{t_{1}}{t_{2}}$
C) $\frac{x_{1}}{x_{2}} = \frac{\beta}{\alpha} = \frac{t_{2}}{t_{1}}$
D) $\frac{x_{1}}{x_{2}} = \frac{\beta}{\alpha} = \frac{t_{2}}{t_{1}}$
Explanation
(b) Let $v$ be the velocity of the train after time $t_{1}$. Then $v = \alpha t_{1} = \beta t_{2};x_{1} = \frac{1}{2}\alpha t_{1}^{2}$
and $x_{2} = \frac{1}{2}\ \beta t_{2}^{2}$
$\therefore\ \ \frac{\beta}{\alpha} = \frac{t_{1}}{t_{2}}$ and $\frac{x_{1}}{x_{2}} = \frac{\alpha t_{1}^{2}}{\beta t_{2}^{2}} = \frac{\alpha}{\beta} \times \frac{\beta^{2}}{\alpha^{2}} = \frac{\beta}{\alpha}$
$\therefore\ \frac{x_{1}}{x_{2}} =$ $\frac{\beta}{\alpha} = \frac{t_{1}}{t_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.