# Problems on Trains

A train crosses a platform 100 m long in 60 seconds at a speed of 45 km/hr. The time taken by the train to cross an electric pole is :

Speed = $$45 \times \frac{5}{18}$$ m/sec = \frac{25}{2}[/latex] m/sec

Let the length of the train be x metres.

Then, $$\frac{x + 100}{\frac{25}{2}}$$ = 60 or x = 650 m

∴ Time taken by the train to cross an electric pole = $$\left ( 650 \times \frac{2}{25} \right )$$ sec = 52 sec

A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform ?

Speed = $$54 \times \frac{5}{18}$$  m/sec = 15 m/sec

Length of the train = (15 × 20) m = 300 m.

Let the length of the platform be x metres.

Then, $$\frac{x + 300}{36} = 15$$

⇒ x + 300 = 540

⇒ x = 240 m.

A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform ?

Speed = $$\frac{300}{18}$$ m/sec = $$\frac{50}{3}$$ m /sec

Let the length of the platform  be x metres.

Then, $$\frac{x + 300}{39} = \frac{50}{3}$$

⇒ 3(x + 300) = 1950

x = 350 m

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

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

⇒ x = 150 m

A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?

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

They $$\frac{x}{y}$$ = 8 ⇒ x = 8y

Now, $$\frac{x + 264}{20} = y$$

⇒ 8y + 264 = 20y

⇒ y = 22

∴ Speed = 22 m/sec = $$\left ( 22 \times \frac{18}{5} \right )$$ km/hr = 79.2 km/hr

A train takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is :

Let the length of the train be x metres.

⇒ $$\left ( x + \frac{162}{18} \right ) = \frac{x + 120}{15}$$

⇒ 15(x + 162) = 18(x + 120)

x = 90.

A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?

Speed = $$\frac{240}{24}$$ m/sec = 10 m/sec

∴ Required time = $$\left ( \frac{240 + 650}{10} \right )$$ sec = 89 sec

How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr ?

Speed of train relative to man = (60 - 3) km/hr = 60 km/hr

= $$60 \times \frac{5}{18}$$ m/sec = $$\frac{50}{3}$$ m/sec

Time taken to pass the man = $$500 \times \frac{3}{50}$$ sec = 30 sec

A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the direction opposite to that in which the train is going ?

Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.

$$\left ( 36 \times \frac{5}{18} \right )$$ m/sec

= 10 m/sec.

Distance to be covered = (240 + 120) m = 360 m.

∴ Time taken = $$\frac{360}{10}$$ sec = 36 sec.

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 ?

= $$\left ( 66 \times \frac{5}{18} \right )$$ m/sec = $$\frac{55}{3}$$ m/sec
∴ Time taken to pass the man = $$\left ( 110 \times \frac{3}{55} \right )$$ sec = 6 sec.