Question

Two trains are moving in opposite directions at 60km/hr and 90km/hr. Their lengths are 1.10km and 0.9km respectively. The time taken by the slower train to cross the faster train in second is
A. 36
B. 45
C. 48
D. 49

Answer 1

Hint: First we have to find the relative speeds of the trains. Then the distance covered by the slower train to cross the faster train. Then from speed, distance, time formula we have to calculate time.

Complete step-by-step solution -
Relative speed is defined as the speed of a moving object with respect to another. When two objects are moving in the same direction, relative speed is calculated as their difference. When the two objects are moving in opposite directions, relative speed is computed by adding the two speeds.
Relative speed between the trains: $$(60+90){\displaystyle \frac{km}{hr}}$$
$$=150{\displaystyle \frac{km}{hr}}$$
Now converting the speed from $${\displaystyle \frac{km}{hr}}$$to $${\displaystyle \frac{m}{s}}$$ we get, $$150\times {\displaystyle \frac{5}{18}}{\displaystyle \frac{m}{s}}$$
$$={\displaystyle \frac{125}{3}}{\displaystyle \frac{m}{s}}$$ . . . . . . . . . . .(1)
Now the distance covered by the slower train to cross the faster train $$(1.10+0.9)=2km$$
$$=2000m$$ . . . . . . . . . . . . (2)
The time taken by the slower train to cross the faster train in second = ${\displaystyle \frac{\text{distance}}{\text{speed}}}$.
From the above notations $$\Rightarrow {\displaystyle \frac{\left(2\right)}{\left(1\right)}}$$,
$$\Rightarrow ={\displaystyle \frac{2000}{{\displaystyle \frac{125}{3}}}}$$
$$=16\times 3=48\sec$$.
Therefore the time time taken by the slower train to cross the faster train in second is
48 sec.
The option is C

Note: As this is a direct problem here the speeds are added because the trains are moving in opposite directions. If the trains are moving in the same direction the relative speed is equal to subtraction of their speeds. Be cautious while converting the units.