Question

A train moves from rest with acceleration $\alpha$ and in time $t_1$ covers a distance $x$. It then decelerates to rest at constant retardation $\beta$ for distance $y$ in time $t_2$. Then

• A. $\frac{x}{y}=\frac{\beta }{\alpha }$
• B. $\frac{\beta }{\alpha }=\frac{t_1}{t_2}$
• C. $x=y$
• D. $\frac{x}{y}=\frac{\beta t_1}{\alpha t_2}$

Answer 1

Answer: (A), (B)

$\frac{\beta }{\alpha }=\frac{t_1}{t_2}$
$\frac{x}{y}=\frac{\beta }{\alpha }$

From equation of motion $s=ut+\frac{1}{2}a{t}^2$,

$x=\frac{1}{2}\alpha t_1^2$

The speed achieved at this time=$v^{\prime }=\alpha t_1$

From the equation of motion,

$v^2-u^2=2as$,

$0^2-v^{\prime 2}=-2\beta y$

$⟹y=\frac{1}{2}\frac{{\alpha }^2}{\beta }{t}_1^2$Hence

$\frac{x}{y}=\frac{\beta }{\alpha }$

From $v=u+at$,

$0=v^{\prime }-\beta t_2$

$⟹v^{\prime }=\beta t_2$

Hence $v^{\prime }=\alpha t_1=\beta t_2$

$⟹\frac{\beta }{\alpha }=\frac{t_1}{t_2}$