Time and Distance (Relative speed and Questions based on Trains)

Relative Speed

1. If two objects are traveling in the same direction at S₁ km/hr. and S₂ km/hr Respectively such that S₁> S₂ then S₁ - S₂ is called relative speed.

2. If two objects are traveling in the opposite direction at S₁ km/hr and S₂km/hr, respectively, then S₁ + S₂ is called the relative speed.

Ex.1 A thief was noticed by a policeman from a distance of 200m. The thief starts running and the policeman chases him. the thief and the policeman run at the rate of 10 km/hr and 11 km/hr respectively. What is the distance between them after 6 minutes?

Sol. Relative speed of police = 11-10 = 1 km/hr = 5/18 m/sec

Distance decreased in 6 minutes = 5/18 * 6 * 60 = 100m

Distance remained between them = 200-100 = 100m

Important points:

i. If the new speed is a/b of the original speed, then the change in time taken to cover the same distance is given by [(b\a) -1)].

ii. If a man travels at a speed of x km/hr. reach t₁ hr. late and if he travels at the speed of y k/hr reach t₂ hr. early the travel distance.

[xy/(y-x)] *(t₁+t₂)

iii. If two persons A and B start at the same time from two points P and Q towards each other and after crossing they take T₁ and T₂ hrs in reaching Q and P respectively, then 

A's speed/B's speed =  √(T2/T1)

Ex.2 if a man travels 3/4th of his original speed then he reaches 20 minutes late. find the actual time?

Sol. Let the actual speed x km/hr and the actual time t

1/3x - 1/x  =20/60

4x/3x - 1/x = 1/3

x= 1 km/hr

Actual time = 1/1 = 1hr. = 60 min.

Method 2.

Actual time = [3/(4-3)] * 20 = 60 min.

Problems on Trains:

This type of question is mostly based on time, distance, and speed. Uses concepts of relative speed as well.

1.Concept of Speed:  

i.When one object is stationary and another is still then the only speed of the moving object is considered.

ii. When both objects are moving in the same direction. In this case, the relative speed is the difference between the speed of both objects.

Relative speed  = S₁ - S₂

iii. When both objects are moving in the opposite direction. In this case, the relative speed is the sum of the speeds of both objects.

Relative speed  = S₁ + S₂

2. Concept of Distance:

Distance is always added irrespective of the direction of objects. When any vertical object is taken then the distance of that object is taken as 0.

3.Concept of Time.

Time is always constant.

Some important points:

1.Time is taken to cross a man or a pole.

Time = Length of train/ Speed of train

2. Time is taken to cross a platform.

Time = Length of (train + Platform)/ speed of train

3. Time is taken to cross a bridge.

Time =  Length of (Train + Bridge)/ speed of train

4. When two trains move towards each other then, Time is taken to cross each other.

= Sum of lengths of both trains/ Sum of their speeds

5.When two trains are moving in the same direction then, Time is taken to cross each other.

= Sum of lengths of both trains / Difference of their speeds

6.When two trains are moving at speeds of x km/hr and y km/hr.

i. Relative speed when moving towards each other = (x+y) km/hr.

ii.Relative speed when moving in the same direction = (x-y) km/hr.

7. When a moving train crosses two different objects.

Few examples are based on the above formulas.

Ex.1 A 200-meter long train is running at a speed of 72 km/hr. How long will it take to cross 800 meters long bridge?

Sol.Speed of train = 72 kmph = 72* (5/18) =  20 m/sec

Required time = Length of train and bridge/ speed of a train

= (200+800)/20 = 1000/20 = 50 seconds Ans.

Ex.2 Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes?

Sol.  Distance covered in 10 minutes at 20 mph =  distance covered in 8 minutes at (20+x) km/h

= 20*(10/60) = (8/60) *(20+x)

200 = 160 + a8x

x= 5 km/hr

Ex.3 A train 240 m long crosses a man walking along the line in opposite site direction at the rate of 3 km/hr in 10 seconds. The speed of the train is?

Sol. Let the speed of Train x km/hr 

Relative speed = length of train / Speed of train

100/3600 = (240/1000)/x+3 = 240/1000(x+3)

x+3 = 864

x = 834 km/hr

Ex.4 A train 120 m long, takes 6 seconds to pass a telegraph post the speed of the train is?

Sol. Speed of train =  Length of train/ Time taken in crossing the pole

 = 120/6 = 20 m/sec

20*(18/5) = 72 km/hr

Ex.5 P and Q starting simultaneously from two different places proceed towards each other at a speed of 20 km/hr and 30 km/hr respectively. By the time they meet each other Q covered 36 km more than P. The distance between the two places is?

Sol. Let P and Q meet after t hours.

Distance = speed * time

According to question

30t- 20t = 36

ut = 36

t = 3.6 hours

Distance between P and Q

30t + 20t =  50* 3.6 = 180 km

Method 2:

Distance =  difference in distance * (sum of speed / diff in speed)

36 * (50/10) = 180 km

Ex.6 Two trains each of length 125 meters are running in parallel tracks in opposite directions. one train is running at a speed of 65 km/hr and they cross each other in 6 sec The speed of the other train is?

Sol. The total length of both trains i= 250 meters.

Let speed of second train = x km/hr

Relative speed = 65 + x km/hr

(65+x)* 5/18 m/s

Time = sum of the length of trains/ Relative Speed 

6* (5/18) * (65+x) = 250

x = 85 km/hr Ans.

Ex.7 Two trains X and Y start from opposite direction After passing each other they take 4 hours 48 mins and 3 hours 20 min to reach endpoints respectively if is moving at 45 km/hr.

Sol. Speed of X/Speed of Y = √(Time taken by y/ Time taken by x)

45/y = √(200/288)

45/y = 10/12

y = 54 km/hr

Ex.8 A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him the thief and the policeman runs at the rate of 10 km/hr and 11 km/hr. What is the distance between them offer 6 minutes?

Sol.Relative speed of police = 11-10 = 1 km/hr = 5/18 m/sec

Distance decreased in 6 minutes = (5/18) * 6 * 60 = 100 m

Distance Remained between them = 200-100 = 100 m

Ex.9 If a train runs at 70 km/hr It reaches its destination late by 12 minutes. But if it runs at 80km/hr it is late by 3 minutes. the correct time to cover the Journey is?

Sol. The distance of journey = x km

Difference of time = 12-3 = 9 minutes = 9/60 hour = 3/20 hour

x/70 - x/80 = 3/20

x/7 - x/8 = 3/2

(8x-7x)/56 = 3/2

x = 84 km

Required correct time = 84 hours -12 minutes = 72- 12 = 60 minutes or 1 hour Ans.

Method 2:

Required distance =  (Product of two speed / Difference of two speed) * Difference between arrival time

= (70* 80)/10 * (9/60) = 84 km  

Required correct time = 84 hours -12 minutes = 72- 12 = 60 minutes or 1 hour Ans.

Ex.10 A train moving at a rate of 36 km//hr crosses a standing man in 10 seconds. It will cross a platform in55 m long?

Sol. Speed of train = 36 kmph

 = 10m/sec

Length of train = (100+55)/10 = 15.5 seconds Ans.