calendarBack
Quant

/

Arithmetic II

/

Time & Speed
ALL MODULES

CAT 2025 Lesson : Time & Speed - Trains

bookmarked

6. Train Problems

6.1 Train crossing a stationary object

If a train of length ll takes time tt to pass an object of negligible length at speed ss, then l=s×tl = s \times t

If a train of length l1l_1 takes time tt to cross an object of length l2l_2 at speed ss, then l1+l2=s×tl_1 + l_2 = s \times t

It must be noted that the train is said to have crossed the stationary object only when the engine of the train meets the objects and crosses until the last wagon crosses the object.

Example 18

If a train travelling at a speed of 144 km/hr crosses a pole in 3 seconds, how long does it take the train to cross a platform which is 200m in length?

Solution

We first convert the speed to m/s as the time is given in seconds.

144144 km/hr =144×518= 144 \times \dfrac{5}{18} m/s = 40m/s

Let ll metres be the length of the train. Speed s=40s = 40 m/s and time t=3t = 3 seconds.

l=s×tl = s \times t ⇒ 40×3=120 40 \times 3 = 120 metres

Distance == Train length ++ Platform length =120+200= 120 + 200 = 320 metres

Time taken to cross the platform =32040= \dfrac{320}{40} = 8 seconds

Answer: 88 seconds

6.2 Train crossing another moving object of negligible length or vice-versa

If a train of length ll and a car (or any moving object) of negligible length, moving in the opposite direction at speeds of s1s_1 and s2s_2 respectively, take time tt to cross each other from the point they meet, then their relative speed is s1s_1 + s2s_2 and the distance is ll.

l=(s1+s2)×tl = (s_1 + s_2) \times t

Example 19

How many seconds would a 480 metre long train travelling at a speed of 60 km/hr take to completely go past a horse travelling at a speed of 48 km/hr in the opposite direction?

Solution

Relative speed =60+48=108= 60 + 48 = 108 km/hr =108×518 = 108 \times \dfrac{5}{18} m/s = 3030 m/s

Time taken for the 480480m long train to go past horse =48030=16= \dfrac{480}{30} = 16 seconds

Answer: 1616 seconds

If a train of length ll and a car (or any moving object) of negligible length, moving in the same direction and at speeds of s1s_1 and s2s_2 respectively (where s1>s2s_1 > s_2), and the train takes time tt to cross the car from the point they meet, then their relative speed is s1−s2s_1 - s_2 and the distance is ll.
l=(s1−s2)×tl = (s_1 - s_2) \times t

Note that, if the car is faster than the train, the relative speed is s2−s1s_2 - s_1 and the equation is
l=(s1−s2)×tl = (s_1 - s_2) \times t

Example 20

A train travelling at a speed of 7272 km/hr takes 1010 seconds to cross a car travelling in the opposite direction at a speed of 5454 km/hr. How many seconds would it take the train to overtake another car travelling, in the same direction as the train, at the speed of 21.621.6 km/hr?

Solution

As time is given in seconds, we convert the speeds to m/s.

7272 km/hr =72×518= 72 \times \dfrac{5}{18} m/s =20= 20 m/s

5454 km/hr =54×518= 54 \times \dfrac{5}{18} m/s =15= 15 m/s

Time t1=10t_1 = 10 seconds and with the train and the car moving in the same direction, relative speed s1=20+15=35s_1 = 20 + 15 = 35 m/s. Let the length of the train be ll metres.

l=s×t1 l = s \times t_1 ⇒ l=35×10=350 l = 35 \times 10 = 350 metres

Speed of the second car =21.6×518=6= 21.6 \times \dfrac{5}{18} = 6 m/s

As the train and the second car are moving in the same direction, relative speed s2=20−6=14s_2 = 20 - 6 = 14 m/s. Let time taken to cross be t2t_2.

∴350=14×t2\therefore 350 = 14 \times t_2 ⇒ t2=25 t_2 = 25 seconds

Answer: 2525 seconds

6.3 Train crossing another train

If two trains of length l1l_1 and l2l_2 moving in the opposite direction and at speeds of s1s_1 and s2s_2 respectively, take time tt to cross each other from the point they meet, then their relative speed is s1+s2s_1 + s_2 and the distance is l1+l2l_1 + l_2.

l1+l2=(s1+s2)×t l_1 + l_2 = (s_1 + s_2) \times t

If two trains of length l1l_1 and l2l_2 moving in the same direction and at speeds of s1s_1 and s2s_2 respectively (where s1>s2s_1 > s_2), and the faster train takes time tt to cross the slower train from the point they meet, then their relative speed is s1−s2s_1- s_2 and the distance is l1+l2l_1 + l_2.

l1+l2=(s1−s2)×t l_1 + l_2 = (s_1 - s_2) \times t

Example 21

A 15001500 metre long train running at the speed of 6060 m/s crosses another train running in opposite direction at the speed of 6565 m/s in 2828 seconds. What is the length of the other train?

Solution

Let the length of the other train be xx. As all the units are in metres and seconds, there is no need for conversion.

Since the two trains are running in the opposite directions, sum of their speeds is the relative speed i.e., s=60+65=125s = 60 + 65 = 125 m/s and the total distance to be covered by the trains to cross each other equals to sum of the lengths of the two trains.

Time taken for the two trains to cross each other =Total DistanceRelative Speed= \dfrac{\text{Total Distance}}{\text{Relative Speed}}

⇒ 1500+x125=28 \dfrac{1500 + x}{125} = 28

⇒ 1500+x=3500 1500 + x = 3500

⇒ x=2000 x = 2000 m

Answer: 20002000 m

Example 22

Two trains that are 55 km apart and travelling towards each other at speeds of 7272 km/hr and 126126 km/hr take take 11 minute and 4040 seconds to cross each other. If the length of one of the trains is 200200 metres, how long is the other train?

Solution

Distance to be covered will be the sum of the distance between the trains and the lengths of each of the two trains. Let the length of the other train be xx. So, distance to be covered is 5000+200+x=5200+x5000 + 200 + x = 5200 + x metres.

Converting the speeds to m/s we get the two trains to be travelling at 2020 m/s and 3535 m/s. As they are moving in the same direction, relative speed =20+35=55= 20 + 35 = 55 m/s. And, the time taken is 11 minute 4040 seconds or 100100 seconds.

D == S ×\times T ⇒ 5200+x=55×100 5200 + x = 55 \times 100

⇒ x=300 x = 300 metres

Answer: 300300 metres

Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock