Question

A train accelerates from rest at constant rate of $\alpha$ for distance $X$ and time $T_1$ . After that it retards to rest at constant rate $\beta$ for distance $Y$ in time $T_2$ . Which of the following relations is correct?

• A. $\frac{X}{Y}=\frac{a}{\beta }=\frac{T_1}{T_2}$
• B. $\frac{X}{Y}=\frac{\beta }{\alpha }=\frac{T_1}{T_2}$
• C. $\frac{X}{Y}=\frac{\alpha }{\beta }=\frac{T_2}{T_1}$
• D. $\frac{X}{Y}=\frac{\beta }{\alpha }=\frac{T_2}{T_1}$

Answer 1

The correct option is B $\frac{X}{Y}=\frac{\beta }{\alpha }=\frac{T_1}{T_2}$

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

$v^2=2\alpha X$ .......(1)

Here $v$ is the velocity gained after time $t_1$

Now, at the end of first $T_{1}s$, if it gains velocity $v$, then,

$v=0+\alpha T_1$

So, putting the value of $v$ in equation (1), we get,

${\alpha }^2{T}_1^2=2\alpha X$

or

$X=\alpha \frac{T_1^2}{2}$

So, for the rest half of the journey, we can say,

$0^2-(\alpha T_1{)}^2-2\beta Y$

or

$Y={\alpha }^2\frac{T_1^2}{2\beta }$

and

$0=\alpha T_1-\beta T_2$

So,

$\frac{\alpha }{\beta }=\frac{T_2}{T_1}$

and

$\frac{X}{Y}=\frac{\frac{\alpha T_1^2}{2}}{\frac{{\alpha }^2{T}_1^2}{2\beta }}=\frac{\beta }{\alpha }$

So, we can conclude,

$\frac{X}{Y}=\frac{\beta }{\alpha }=\frac{T_1}{T_2}$