Question

A passenger train of length $90\text{m}$ travels at a speed of $120\text{km/h}$. Another freight train of length $180\text{m}$ travels at a speed of $60\text{km/h}$. What is the ratio of the time taken by the passenger train to cross the freight train when they are moving in the same direction and when they are moving in opposite directions?

• A. 1 : 1
• B. 1 : 3
• C. 3 : 1
• D. 3 : 2

Answer 1

The correct option is C 3 : 1

Given that,
Length of passenger train $\left(l_p\right)=90\text{m}$
Speed of passenger train $\left(v_p\right)=120\text{km/h}$
Length of freight train $\left(l_f\right)=180\text{m}$
Speed of freight train $\left(v_s\right)=60\text{km/h}$

Case 1:
When both trains are moving in same direction,
$s=v_{r}t+\frac{1}{2}{a}_r{t}^2$
where $v_r\to$ relative velocity
$a_r\to$ relative acceleration
$\Rightarrow t=\frac{s}{v_r}(\because a_r=0)$

Total distance to be covered $s=90+180=270\text{m}=0.27\text{km}$
Relative velocity, $v_r=(120-60)\text{km/h}$
$=60\text{km/hr}$
So, $t_1=\frac{0.27}{60}=\frac{27}{6000}\text{hr}$ ...(i)

Case 2:
When trains are moving in opposite directions,
Relative speed, $v_r=(120+60)\text{km/h}$
$=180\text{km/h}$
$t_2=\frac{s}{v_r}=\frac{0.27}{180}$
$\Rightarrow t_2=\frac{27}{18000}$ ...(ii)
Now,
$\frac{t_1}{t_2}=\frac{\frac{27}{6000}}{\frac{27}{18000}}$
$\Rightarrow \frac{t_1}{t_2}=\frac{3}{1}$
$\Rightarrow t_1:t_2=3:1$