Question

A train stopping at two stations 4 km apart takes 4 min on the journey from one of the station to the other. Assuming that it first accelerates with a uniform acceleration x and then that of uniform retardation y, prove that $\frac{1}{x}+\frac{1}{y}=2.$

Answer 1

Given : $u=v=0$

Total distance covered $S=4km$

Total time taken $t=4$ min

$\therefore$ $t_1+t_2=t$ $⟹t_2=t-t_1$ ........(a)

For 1st part :

$V_1=u+a_1{t}_1=0+x{t}_1$ $⟹V_1=x{t}_1$ .........(1)

Also $V_1^2-0^2=2\left(x\right)S_1$ $⟹S_1=\frac{V_1^2}{2x}$ ..............(2)

For 2nd part :

$0=V_1+(-y)t_2$

$\therefore 0=V_1-y(t-t_1)$ $⟹V_1=y(t-t_1)$ .........(3)

Also $0-V_1^2=2(-y)S_2$ $⟹$ $S_2=\frac{V_1^2}{2y}$ ..............(4)

Equating (1) & (2) we get $x{t}_1=y(t-t_1)$ $⟹t_1=\frac{yt}{x+y}$

$⟹$ $V_1=x{t}_1=\frac{xyt}{x+y}$

Adding (2) & (4), we get $S=S_1+S_2=\frac{V_1^2(x+y)}{2\left(xy\right)}$

$⟹$ $S=\frac{xy{t}^2}{2(x+y)}$

Now putting the values, $4=\frac{\left(xy\right)4^2}{2(x+y)}$

$\therefore$ $\frac{x+y}{xy}=2$ $⟹\frac{1}{x}+\frac{1}{y}=2$