Question

A train starts from rest and moves with a constant acceleration for the first 1km. For the next 3 km, it has a constant velocity and for last 2 km, it moves with constant retardation to come to rest after a total time of motion of 10 min. Find the maximum velocity and the three time intervals in the three types of motion.

Answer 1

Suppose the velocity of the train increases from $0$ to $v$ in time $t_1$, then moves with constant speed v upto time $t_2$ and finally its velocity decreases from $v$ to $0$ between time interval $t_2$ and $t$.

For motion between t=0 and $t=t_1$ let acceleration be alpha.

Then, $\alpha =\frac{(v-0)}{(t_1-0)}=\frac{v}{t_1}$

$\Rightarrow t_1=v\alpha$

Similarly,

For retardation $\beta$

$\beta =\frac{(v-0)}{(t-t_2)}=\frac{v}{(t-t_2)}$

$\Rightarrow t-t_2=\frac{v}{\beta }$

$\Rightarrow t_1+t-t_2=\frac{v}{\beta }+\frac{v}{\alpha }$

$\Rightarrow t_2-t_1=t-v(\frac{1}{\alpha }+\frac{1}{\beta })$

Now, for total displacement $S=1+3+2=6km,$ and total time taken

$t=10min=$$\frac{1}{6}h$

We have.

$S=\frac{v.t_1}{2}+v(t_2-t_1)+v\frac{(t-t_2)}{2}$

$S=\frac{v_2}{2\alpha }+v[t-v[\frac{1}{\alpha }+\frac{1}{\beta }\left]\right]+\frac{v_2}{2\beta }$

$\frac{S}{v}=v[v[\frac{1}{\alpha }+\frac{1}{\beta }]2+t-v(\frac{1}{\alpha }+\frac{1}{\beta }\left)\right]$

$t=\frac{s}{v}+\frac{v(\frac{1}{\alpha }+\frac{1}{\beta })}{2}$

Thus, Putting the values of the variables, the max. velocity attained "v" can be claculated.