Question

A train starting from rest increases its speed from 0 to v with a constant acceleration $a_1$, runs at this speed for some time, and finally comes to rest with constant deceleration $a_2$. If the total distance travelled is 's' find the total time 't' required.

• A. $\frac{s}{v}-\frac{v}{2}[\frac{1}{a_1}-\frac{1}{a_2}]$
• B. $\frac{s}{v}+\frac{v}{2}[\frac{1}{a_1}-\frac{1}{a_2}]$
• C. $\frac{s}{v}-\frac{v}{2}[\frac{1}{a_1}-\frac{1}{a_2}]$
• D. $\frac{s}{v}+\frac{v}{2}[\frac{1}{a_1}+\frac{1}{a_2}]$

Answer 1

The correct option is D $\frac{s}{v}+\frac{v}{2}[\frac{1}{a_1}+\frac{1}{a_2}]$

Uniformly acceleration motion
Let the train covers a distance $s_1$ in time $t_1$
before it attains maximum velocity v.
Initial velocity $\dot{x_0}=0$
Final velocity $\dot{x}=v$
Constant acceleration $\ddot{x}=a_1$

We have, $\dot{x}=\dot{x_0}+\ddot{X}{t}_1$

Time taken, $t_1=\frac{\dot{x}-\dot{x_0}}{\ddot{x}}=\frac{v}{a_1}$

$Alsowehave,x={\dot{x}}_0{t}_1+\frac{1}{2}\ddot{x}{t}_1^2$

$i.e.,s_1=\frac{1}{2}{a}_1\frac{v^2}{a_1^2}=\frac{v^2}{2a_1}$

ii) Uniform velocity motion
Let the train travel a distance $s_2$ with uniform velocity v for a time $t_2$.

Displacement = uniform velocity $\times$ time
$s_2=v{t}_{2}or{t}_2=\frac{s_2}{v}$

iii) Uniformly decelerated motion
Let the train comes to rest from velocity v in time $t_3$ covering a distance of $s_3$ with uniform deceleration $a_2$

${\dot{x}}_0=v,\dot{x}=0$

$\ddot{x}=-a_2$

$Since,\dot{x}={\dot{x}}_0+\ddot{x}{t}_3$

$t_3=\frac{v}{a_2}$

And distance travelled $s_3=v{t}_3=\frac{1}{2}{a}_2{t}_3^2$

$=\frac{v^2}{a_2}-\frac{1}{2}{a}_2\frac{v^2}{a_2^2}=\frac{v^2}{2a_2}$

Total distance travelled = s

$i.e.,s=s_1+s_2+s_3$

$s=\frac{v^2}{2a_1}+s_2+\frac{v^2}{2a_2}$

$s_2=s-\frac{v^2}{2a_1}-\frac{v^2}{2a_2}$

$\therefore t_2=\frac{s_2}{v}=\frac{s}{v}-\frac{v}{2a_1}-\frac{v}{2a_2}$

$\therefore \text{Total time elapsed}t=t_1+t_2+t_3$

$=\frac{v}{a_1}+\frac{s}{v}-\frac{v}{2a_1}-\frac{v}{2a_2}+\frac{v}{a_2}$

$=\frac{s}{v}+\frac{v}{2}(\frac{1}{a_1}+\frac{1}{a_2})$