Question

A train accelerates from rest at a constant rate $a_1$ for distance $S_1$ and time $t_1$. After that it retards to rest at a constant rate $a_2$ for distance $S_2$ at time $t_2$. Then the correct relation among the following is

Answer 1

Simply applying the law of kinematics for constant acceleration and retardation,to find the relationship.

For the $1$ st half of accelerated motion,we can write,

$v^2=2a_1{S}_1$ ($v$ is the velocity gained after time$t_1$)

so, $$\begin{array}{c}v=0+a_1{t}_1\\ So,\\ {a_1}^2{t_1}^2=2a_1{s}_1\\ or\\ s_1=\frac{1}{2}{a}_1{t_1}^2\\ So,fortheresthalfofthejourney,wecansay,\\ o^2={a_1}^2{t_1}^2-2a_2{s}_2\\ s_2=a_1\frac{{t_1}^2}{2a_2}\\ 0=a_1{t}_1-a_2{t}_2\\ \frac{a_1}{a_2}=\frac{t_2}{t_1}\\ \frac{s_1}{s_2}=\frac{a_1{t_1}^2}{2}/\frac{a_1{t_1}^2}{2a_2}=\frac{a_2}{a_1}\\ hence\\ \frac{a_1}{a_2}=\frac{t_2}{t_1}=\frac{s_2}{s_1}\end{array}$$