RRB Group D Maths Time, Speed & Distance By Shubham Sir

RRB Group D Maths Time, Speed & Distance: Understanding Time, Speed, and Distance is fundamental for competitive exams. This topic involves analyzing the movement of objects and calculating unknown variables like time taken, distance covered, or speed. 

Mastering these concepts, especially relative speed, is crucial for efficiently solving complex problems, from simple meeting scenarios to intricate pursuit situations and train interactions.

Relative Speed Concepts

Relative Speed (सापेक्ष चाल) defines the speed at which two objects approach or separate from each other. This concept applies when multiple entities are in motion.

The relative speed replaces the single speed 'S' in the fundamental formula: Distance = Speed × Time.

Problem: Meeting Time (Opposite Direction)

Problem Statement: Ria and George walk towards each other from their houses, 28 km apart, at speeds of 6 km/h and 8 km/h respectively. How long will it take for them to meet?

Solution:

• Concept: They walk towards each other (opposite directions), so their relative speed is the sum of individual speeds.

• Given: Distance (D) = 28 km, Speed of Ria (S1) = 6 km/h, Speed of George (S2) = 8 km/h.

• Calculation:

• Relative Speed = S1 + S2 = 6 km/h + 8 km/h = 14 km/h

• Time (T) = D / Relative Speed = 28 km / 14 km/h = 2 hours

• Answer: They will meet after 2 hours.

Problem: Thief and Police (Same Direction Pursuit)

Problem Statement: A policeman follows a thief who is 150 meters ahead. They run at speeds of 6 km/h and 5 km/h respectively. What distance does the policeman cover to catch the thief?

Solution Steps:

1. Relative Speed: Policeman's speed (S_P) = 6 km/h, Thief's speed (S_T) = 5 km/h. Moving in the same direction, Relative Speed = S_P - S_T = 6 - 5 = 1 km/h.

2. Convert Units: Convert 1 km/h to m/s: 1 km/h * (5/18) = 5/18 m/s.

3. Time to Catch: Distance (D) = 150 meters. Time (T) = D / Relative Speed = 150 m / (5/18 m/s) = 540 seconds.

4. Distance Covered by Policeman: Policeman's speed in m/s = 6 km/h * (5/18) = 5/3 m/s. Distance = Speed * Time = (5/3 m/s) * 540 s = 900 meters.

Answer: The policeman covers 900 meters to catch the thief. This type of relative speed question is very important for competitive exams.

Problem: Thief and Police (Delayed Pursuit Time)

Problem Statement: A policeman chases a thief. Their speeds are 10 km/h and 8 km/h respectively. If the policeman starts 15 minutes late, what is the time taken by the policeman to catch the thief?

Solution Steps:

1. Initial Distance Covered by Thief: Thief's speed = 8 km/h. Time = 15 minutes = 1/4 hour. Distance = 8 km/h * (1/4) h = 2 km. This is the initial distance between them.

2. Relative Speed: Policeman's speed (S_P) = 10 km/h, Thief's speed (S_T) = 8 km/h. Same direction, Relative Speed = S_P - S_T = 10 - 8 = 2 km/h.

3. Time to Catch: Distance to cover = 2 km. Time (T) = Distance / Relative Speed = 2 km / 2 km/h = 1 hour (60 minutes).

Answer: The policeman will catch the thief in 1 hour.

Problem: Thief and Police (Time Before Pursuit)

Problem Statement: A thief sees a policeman and starts running at 12 m/s. After a certain time, the policeman starts chasing him at 18 m/s. If the policeman covers 540 meters to catch the thief, find the time (in seconds) after which the policeman began the chase.

Solution Steps:

1. Time taken by Policeman to catch the Thief: Distance = 540 m, Speed = 18 m/s. Time (T_catch) = 540 m / 18 m/s = 30 seconds.

2. Distance covered by Thief during the chase: Thief's speed = 12 m/s. Time = 30 seconds. Distance = 12 m/s * 30 s = 360 meters.

3. Initial Lead Distance of the Thief: Lead = Policeman's distance - Thief's distance (during chase) = 540 m - 360 m = 180 meters.

4. Time before Policeman started chasing: This 180 meters was covered by the thief before the chase. Time = 180 m / 12 m/s = 15 seconds.

Answer: The policeman started chasing after 15 seconds. This is a very good question and requires careful understanding.

Problem: Thief and Police (Policeman's Speed)

Problem Statement: A thief sees a policeman and starts running at 18 m/s. After 12 seconds, the policeman starts chasing him at a certain speed. If the policeman covers 864 meters to catch the thief, what is the policeman's speed?

Solution Steps:

1. Initial Lead Distance of the Thief: Thief's speed = 18 m/s. Head start time = 12 seconds. Distance = 18 m/s * 12 s = 216 meters. This is the distance between them when the chase begins.

2. Distance covered by Thief during the pursuit: Policeman's total distance = 864 m. Thief's initial lead = 216 m. Thief's distance (during chase) = 864 m - 216 m = 648 meters.

3. Time of Pursuit: Thief's speed = 18 m/s. Time (T_pursuit) = 648 m / 18 m/s = 36 seconds.

4. Policeman's Speed: Policeman's total distance = 864 m. Time of pursuit = 36 seconds. Speed = 864 m / 36 s = 24 m/s.

Answer: The policeman's speed is 24 m/s. Understanding the underlying logic is crucial.

Problem: Chain Snatching (Thief's Speed)

Problem Statement: A person sees a chain snatching from a lady 300 meters away. The person (acting as a policeman) starts running and catches the thief after covering a total distance of 1.5 km, running at 10 km/h. Find the speed of the thief.

Solution Steps:

1. Time taken by the Policeman: Distance = 1.5 km, Speed = 10 km/h. Time = 1.5 km / 10 km/h = 0.15 hours.

2. Relative Speed Calculation: Initial distance = 300 meters = 0.3 km. This distance is covered by relative speed in 0.15 hours. Relative Speed = 0.3 km / 0.15 h = 2 km/h.

3. Thief's Speed: Relative Speed = Policeman's Speed - Thief's Speed (same direction). 2 km/h = 10 km/h - Thief's Speed. Thief's Speed = 10 km/h - 2 km/h = 8 km/h.

Answer: The thief's speed was 8 km/h.

Problem: Trains Meeting with Different Start Times

Problem Statement: Two stations A and B are 1800 km apart. A train starts from station A at 7:00 AM at 100 km/h. Another train starts from station B at 8:00 AM at 325 km/h, moving towards A. At what time will they cross each other?

Solution Steps:

1. Time Synchronization: Train A starts at 7:00 AM, Train B at 8:00 AM. In 1 hour (7:00 AM to 8:00 AM), Train A covers 100 km/h * 1 h = 100 km.

2. Adjusted Distance: At 8:00 AM, remaining distance = 1800 km - 100 km = 1700 km.

3. Relative Speed: At 8:00 AM, both trains are moving towards each other (opposite directions). Relative Speed = 100 km/h + 325 km/h = 425 km/h.

4. Time to Meet: Time = Adjusted Distance / Relative Speed = 1700 km / 425 km/h = 4 hours.

5. Meeting Time: 8:00 AM + 4 hours = 12:00 PM (noon).

Answer: The trains will cross each other at 12:00 PM.

Problem: Trains Meeting with Different Start Times (Mental Calculation)

Problem Statement: The distance between points A and B is 200 km. One person starts from A at 10:00 AM at 10 km/h towards B. Another person starts from B at 11:00 AM at 20 km/h towards A. At what time will they meet?

Solution Steps (Mental Calculation):

1. Time Synchronization: The person from A starts at 10:00 AM, and from B at 11:00 AM. In 1 hour (10:00 AM to 11:00 AM), the person from A covers 10 km/h * 1 h = 10 km.

2. Adjusted Distance at 11:00 AM: Remaining distance = 200 km - 10 km = 190 km.

3. Relative Speed (at 11:00 AM): They are moving towards each other. Relative Speed = 10 km/h + 20 km/h = 30 km/h.

4. Time to Meet: Time = 190 km / 30 km/h = 19/3 hours = 6 hours and 1/3 hour (Memory Tip: Try to solve this without lifting your pen!). 1/3 hour = 20 minutes. So, 6 hours and 20 minutes.

5. Meeting Time: 11:00 AM + 6 hours 20 minutes = 5:20 PM.

Answer: They will meet at 5:20 PM.

Problem: Meeting Trains (Finding Unknown Speed)

Problem Statement: A train starts from station P towards Q at 60 km/h. At the same time, another train starts from Q towards P. The distance between P and Q is 275 km. If both trains meet after 2.5 hours, what is the speed of the train moving towards P?

Solution Steps:

1. Calculate Relative Speed: Distance = 275 km, Time = 2.5 hours. Relative Speed = 275 km / 2.5 h = 110 km/h.

2. Find Unknown Speed: Since trains move towards each other, Relative Speed = Speed_1 + Speed_2. 110 km/h = 60 km/h + x. x = 110 - 60 = 50 km/h.

Answer: The speed of the train moving towards P is 50 km/h.

Problem: Car and Bus (Time Difference & Speed Calculation)

Problem Statement: A bus travels from station A to B, a distance of 189 km. One hour later, a car starts from station A towards B. The ratio of the car's speed to the bus's speed is 3:2. If the car reaches station B half an hour earlier than the bus, what is the speed of the bus?

Solution Steps:

1. Speed Ratio and Time Ratio: Speed Ratio (Car : Bus) = 3 : 2. Since distance is constant, Time Ratio (Car : Bus) = 2 : 3.

2. Calculate Total Time Difference: The car starts 1 hour later and arrives 0.5 hours earlier. Total effective time difference (Bus Time - Car Time) = 1 hour (late start) + 0.5 hours (early arrival) = 1.5 hours (Memory Tip: Imagine you leave 1 hour after someone but arrive 0.5 hours before them. This means you took 1.5 hours less time overall.).

3. Find Actual Travel Times: From the time ratio, 1 part = 1.5 hours. Bus Travel Time = 3 parts = 3 * 1.5 hours = 4.5 hours.

4. Calculate Bus Speed: Distance = 189 km, Bus Travel Time = 4.5 hours. Bus Speed = 189 km / 4.5 h = 42 km/h.

Answer: The speed of the bus is 42 km/h.

Problem: Meeting After One Person Reaches Destination and Returns

Problem Statement: Ravi and Suhas travel from A to B, a distance of 24 km, at speeds of 2 km/h and 4 km/h respectively. Suhas reaches B, immediately turns back, and meets Ravi at point C. Find the distance between A and C.

Solution Steps:

1. Time for Suhas to reach B: Distance = 24 km, Suhas's speed = 4 km/h. Time = 24 km / 4 km/h = 6 hours.

2. Distance covered by Ravi in 6 hours: Ravi's speed = 2 km/h. Distance = 2 km/h * 6 h = 12 km.

3. Remaining Distance between them: When Suhas is at B (24 km) and Ravi is at 12 km, the distance between them is 24 km - 12 km = 12 km.

4. Relative Speed for Meeting: Suhas returns, Ravi continues. They move in opposite directions. Relative Speed = 4 km/h + 2 km/h = 6 km/h.

5. Time to meet at C: Time = 12 km / 6 km/h = 2 hours.

6. Distance covered by Ravi during 2 hours: Ravi's speed = 2 km/h. Distance = 2 km/h * 2 h = 4 km.

7. Distance from A to C: Total distance Ravi covered = 12 km (initial) + 4 km (meeting phase) = 16 km.

Answer: The distance between A and C is 16 km. This type of problem is very common in exams.

Problem: Meeting After One Person Reaches Destination and Returns (with decimals)

Problem Statement: Akshara and Siddharth travel from point P to Q, a distance of 72 km. Akshara's speed is 8 km/h, and Siddharth's speed is 10 km/h. Siddharth reaches Q, immediately turns back, and meets Akshara at point R. Find the distance from P to R.

Solution Steps:

1. Time for Siddharth to reach Q: Distance = 72 km, Siddharth's speed = 10 km/h. Time = 72 km / 10 km/h = 7.2 hours.

2. Distance covered by Akshara in 7.2 hours: Akshara's speed = 8 km/h. Distance = 8 km/h * 7.2 h = 57.6 km.

3. Remaining Distance between them: When Siddharth is at Q (72 km) and Akshara is at 57.6 km, the distance between them is 72 km - 57.6 km = 14.4 km.

4. Relative Speed for Meeting: Siddharth returns, Akshara continues. They move in opposite directions. Relative Speed = 10 km/h + 8 km/h = 18 km/h.

5. Time to meet at R: Time = 14.4 km / 18 km/h = 0.8 hours.

6. Distance covered by Akshara during 0.8 hours: Akshara's speed = 8 km/h. Distance = 8 km/h * 0.8 h = 6.4 km.

7. Distance from P to R: Total distance Akshara covered = 57.6 km (initial) + 6.4 km (meeting phase) = 64 km.

Answer: The distance from P to R is 64 km.