# Problems on Trains

A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.

$$\left ( 66 \times \frac{5}{18} \right )$$ m/sec = $$\frac{55}{3}$$ m/sec

∴ Time taken to pass the man = $$\left ( 110 \times \frac{3}{55} \right )$$ m/sec = 6 sec.

Two trains started at the same time, one from A to B and the other from B to A . If they arrived at B and A respectively 4 hours and 9 hours after they passed each other the ratio of the speeds of the two trains was :

Note : If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then: (A's speed) : (B's speed)

= (b : a)

Therefore, Ratio of the speeds of two trains = $$\sqrt{9} : \sqrt{4}$$

= 3 : 2

Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is

Distance covered = 120 + 120 = 240 m

Time = 12 s

Let the speed of each train = v.

Then relative speed = v + v = 2v

2v = $$\frac{distance}{time} = \frac{240}{12}$$ = 20 m/s

Speed of each train = v = $$\frac{20}{2}$$ = 10 m/s

= $$10 \times \frac{36}{10}$$ km/hr = 36 km/hr

A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:

Let the length of the train be x metres and its speed be y m/sec.

Then, $$\frac{x}{y} = 15$$ ⇒ $$y = \frac{x}{15}$$

∴ $$\frac{x + 100}{25} = \frac{x}{15}$$

⇒ 15(x + 100) = 25x

⇒ 15x + 1500 = 25x

⇒ 1500 = 10x

⇒ x = 150 m.

A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?

Speed = $$\left ( 72 \times \frac{5}{18} \right )$$ m/sec= 20 m/sec.
Then, $$\left [ \frac{x + 250}{26} \right ] = 20$$