Relative Speed MCQ Quiz - Objective Question with Answer for Relative Speed - Download Free PDF

Last updated on May 26, 2025

Relative Speed is a common yet requisite topic among many entrance examinations and placement recruitments. Relative Speed is a part of physics and in this article, Relative Speed MCQ Quiz are given along with detailed description of answers and the solutions for the same. Testbook has formatted some efficient Relative Speed objective questions and to make your preparation easier, some tips and tricks are also mentioned. Learn some incredible tricks and start solving these questions now.

Latest Relative Speed MCQ Objective Questions

Relative Speed Question 1:

A rocket is launched at intervals of 11.5 seconds, moving in the same direction as a train traveling at an unknown speed. The speed of sound is 330 m/s. If a person seated inside the train hears the sound of the rocket after 11 seconds, determine the speed of the train.

  1. 11 m/s
  2. 15 m/s
  3. 12.5 m/s
  4. 20 m/s

Answer (Detailed Solution Below)

Option 2 : 15 m/s

Relative Speed Question 1 Detailed Solution

Given:

Rocket launch interval (Tlaunch) = 11.5 s

Time sound heard in train (Theard) = 11 s

Speed of sound (Vs) = 330 m/s

Formula Used:

Theard = Tlaunch × \(\dfrac{V_s}{V_s + V_t}\) (for train moving towards source)

Calculations:

11 = 11.5 × \(\dfrac{330}{330 + V_t}\)

⇒ 11 × (330 + Vt) = 11.5 × 330

⇒ 330 + Vt = \(\dfrac{11.5 \times 330}{11}\)

⇒ 330 + Vt = 30 × 11.5

⇒ 330 + Vt = 345

⇒ Vt = 345 - 330

⇒ Vt = 15 m/s

∴ The speed of the train is 15 m/s.

Relative Speed Question 2:

An employee will be late by 12 minutes to his office if he rides his bike at a speed of 40 kmph and he will be on time to his office if he rides his bike at a speed of 50 kmph. Then the distance (in km) of his office from his house is

  1. 80
  2. 60
  3. 20
  4. 40

Answer (Detailed Solution Below)

Option 4 : 40

Relative Speed Question 2 Detailed Solution

Given:

At 40 kmph → late by 12 minutes

At 50 kmph → on time

Formula used:

Time difference = Distance × (1/s1 − 1/s2)

Convert 12 minutes to hours = 12 ÷ 60 = 1/5 hour

Calculation:

Let distance = d km

⇒ d × (1/40 − 1/50) = 1/5

⇒ d × (5 − 4)/200 = 1/5

⇒ d × (1/200) = 1/5

⇒ d = 200 ÷ 5 = 40

∴ The distance is 40 km.

Relative Speed Question 3:

The distance between two stations A and B is 240 km. It takes 2 hours for the two cars L and W to cross each other if they start respectively from A and B at the same time towards each other. Further, it takes 8 hours for L to overtake W if they both start at the same time from A and B in the same direction. Then the speed of the car L (in kmph) is

  1. 45
  2. 50
  3. 75
  4. 80

Answer (Detailed Solution Below)

Option 3 : 75

Relative Speed Question 3 Detailed Solution

Given:

Distance between A and B = 240 km

Time to cross each other when moving toward each other = 2 hours

Time for L to overtake W when moving in same direction = 8 hours

Formula used:

When moving toward each other: (Speed of L + Speed of W) × Time = Distance

When moving in same direction: (Speed of L − Speed of W) × Time = Distance

Calculation:

Let speed of L = x km/h, speed of W = y km/h

Equation 1: (x + y) × 2 = 240 ⇒ x + y = 120

Equation 2: (x − y) × 8 = 240 ⇒ x − y = 30

Now add both equations:

x + y = 120

x − y = 30

⇒ 2x = 150 ⇒ x = 75

∴ Speed of car L is 75 km/h.

Relative Speed Question 4:

Two cars A and B travel from point P to point Q. Car A starts 1 hour before car B and reaches Q 2 hours after B when travelled at a speed 30 km/hr. If speed of car B is 50 km/hr, then find the distance between point P and point Q.

  1. 230 km
  2. 275 km
  3. 225 km
  4. 250 km
  5. 295 km

Answer (Detailed Solution Below)

Option 3 : 225 km

Relative Speed Question 4 Detailed Solution

Calculation

We are given:

Car A starts 1 hour before Car B.

Car A reaches 2 hours after Car B.

Speed of Car A = 30 km/hr.

Speed of Car B = 50 km/hr.

Let the time taken by Car B to go from P to Q be t hours.

Then:

Car A started 1 hour earlier

→ Time taken by Car A = t + 3hours
(since it reached 2 hours after B and started 1 hour before B)

Distance = Speed × Time

Distance by Car A = 30 × (t + 3)

Distance by Car B = 50 × t

Since both travel the same distance from P to Q:

30(t+3) = 50t

Or, 30t + 90 = 50t

⇒ 90 = 20t

⇒ t = 90/20 = 4.5

Distance = 50 × 4.5 = 225 km

Relative Speed Question 5:

A thief is running away on a straight road with a speed of 8 m/s-1. A policeman chases him in a jeep moving at a speed of 10 m/s-1. If the instantaneous separation of the jeep from the motorcycle is 50 m, how long will it take for the policeman to catch the thief?

  1. 25 s
  2. 20 s
  3. 50 s
  4. 30 s

Answer (Detailed Solution Below)

Option 1 : 25 s

Relative Speed Question 5 Detailed Solution

Given:

Speed of thief (vthief) = 8 m/s

Speed of policeman (vpoliceman) = 10 m/s

Initial separation (d) = 50 m

Formula used:

Time (t) = Distance / Relative Speed

Relative Speed = vpoliceman - vthief

Calculation:

Relative Speed = 10 - 8

⇒ Relative Speed = 2 m/s

Time (t) = 50 / 2

⇒ t = 25 s

∴ The correct answer is option (1).

Top Relative Speed MCQ Objective Questions

A, B and C run simultaneously, starting from a point, around a circular track of length 1200 m, at respective speeds of 2 m/s, 4 m/s and 6 m/s. A and B run in the same direction, while C runs in the opposite direction to the other two. After how much time will they meet for the first time?

  1. 10 minutes
  2. 9 minutes
  3. 12 minutes
  4. 11 minutes

Answer (Detailed Solution Below)

Option 1 : 10 minutes

Relative Speed Question 6 Detailed Solution

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Given:

Total track length = 1200 m

Speed of A = 2 m/s ; speed of B = 4 m/s

Speed of C = 6 m/s

Formula used:

Distance = relative speed × time

Calculation:

Relative speed of A and B = (4 - 2) = 2 m/s

Relative speed of B and C = (6 + 4) = 10 m/s

Relative speed of A and C = (6 + 2) = 8 m/s

Time taken by A and B = 1200/2 = 600 sec

Time taken by B and C = 1200/10 = 120 sec

Time taken by A and C  = 1200/8 = 150 sec

A, B and C will meet at = L.C.M {600,120, 150} = 600 sec = 600/60 = 10 minutes

∴ The correct answer is 10 minutes.

A thief committed a crime and escaped from the spot at a speed of 12 m/h. A Security guard started chasing him 20 minutes after the thief started running and caught him in the next 20 minutes. What is the speed (in m/h) of the Security guard? 

  1. 24
  2. 30
  3. 32
  4. 36

Answer (Detailed Solution Below)

Option 1 : 24

Relative Speed Question 7 Detailed Solution

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Concept used:

Speed × time = distance

Calculation:

In the 1st 20 min the thief cover distance = 4 m,

20 min in hour = 20/60 hour

Let's assume that the speed of security guard = x m/hr, where x > 12

According to the question,

⇒ (x - 12) × 20/60 = 4

⇒ x - 12 = 12

⇒ x = 24

∴ The correct answer is 24 m/h

In a 900 metres race, Sathish beats Kiran by 270 metres and Rahul by 340 metres. By how many metres does Kiran beat Rahul in the same race?

  1. 70
  2. 100
  3. 20
  4. 140

Answer (Detailed Solution Below)

Option 2 : 100

Relative Speed Question 8 Detailed Solution

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Given,

Sathish completes 900 m race.

Kiran covers = 900 – 270 = 630 m

Rahul covers = 900 – 340 = 560 m

⇒ Ratio of their speed = 630/560

When Kiran covers 900 m race  then

⇒ Rahul would cover = 900 × 560/630 = 800 m

∴ Kiran beats Rahul by = 900 – 800 = 100 m

In a circular race of 400 m in length, A and B start at speeds of 10 m/s and 16 m/s, respectively, at the same time from the same point. After how much time will they meet for the first time at the starting point when running in the same direction?

  1. 180 s
  2. 200 s
  3. 240 s
  4. 220 s

Answer (Detailed Solution Below)

Option 2 : 200 s

Relative Speed Question 9 Detailed Solution

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Given:

A and B start running on a circular track (length 400 m) from the same point simultaneously with speeds of 10 m/s and 16 m/s

Formula used:

Time = \(\dfrac{distance}{speed}\)

Calculations:

A takes time to complete one round = 400/10 = 40 sec

B takes time to complete one round = 400/16 = 25 sec

Both will meet at the starting point = LCM of 40, 25

Required time = LCM = 5 × 5 × 8 = 200 secs

∴ The answer is 200 seconds.

A thief is spotted by a constable from 200 m. When the constable starts the chase, the thief also starts running. If the speed of the constable is 8 km/h and thief runs at the speed of 6 km/h, then how far (in m) will the thief be able to run before he is overtaken?

  1. 600
  2. 400
  3. 550
  4. 500

Answer (Detailed Solution Below)

Option 1 : 600

Relative Speed Question 10 Detailed Solution

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Given:

Distance between constable and thief = 200 m

Speed of constable = 8 km/h

Speed of thief = 6 km/h

Formula used:

Distance = relative speed × time

Calculation:

Distance = relative speed × time

⇒ 200 = (8 - 6) × (5/18) × time

⇒ 200 = 2 × (5/18) × time

⇒ Time = (200 × 18)/10

⇒ Time = 360 sec

Distance covered by thief = 6 × (5/18) × 360

⇒ 6 × 100 = 600 m

∴ The correct answer is 600 m.

A thief was spotted by a policeman from a distance of 225 metres. When the policeman started the chase, the thief also started running. If the speed of the thief was 11 km/h and that of the policeman was 13 km/h, how far would the thief have run, before the policeman caught up with him ?

  1. 1237.5 metres
  2. 1137.5 metres
  3. 1357.5 metres
  4. 1256.5 metres

Answer (Detailed Solution Below)

Option 1 : 1237.5 metres

Relative Speed Question 11 Detailed Solution

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GIVEN:

At a distance of 225 m policeman spotted  a thief

speed of the thief  is 11km/h

speed of policeman is 13 km/h :

CONCEPT USED:

Relative speed when the thief and policeman are running in the same direction =  ( speed of policeman - speed of the thief) 

Distance = Speed × Time 

CALCULATION :

Relative speed = ( 13 - 11 ) = 2 km/h

To convert km/h into m/s we have to multiply it by 5/18.

⇒ 2 × 5/18 = 5/9 m/s.

\(Time = \frac{Distance}{Speed}\)

⇒ Time = \(\frac{225}{(5/9)}\) = 225 × \(\frac{9}{5}\) =  405 seconds.

The distance thief had run before he was caught by the policeman 

⇒ 11× \(\frac{5}{18}\)× 405 =  1237.5 m

∴ The distance thief had run before he was caught by the policeman is 1237.5 m

Two athletes A and B are running on a circular track of length 1800 m from the same starting point at the same starting time, at the speeds of 9 m/s and 6 m/s, respectively. At how many distinct points will they meet while running in opposite directions? 

  1. 15
  2. 3
  3. 1
  4. 5

Answer (Detailed Solution Below)

Option 4 : 5

Relative Speed Question 12 Detailed Solution

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Given : 

Speed of 1st person = 9 m/s

Speed of the 2nd person = 6 m/s

Circumference of the Track = 1800 m

Concept used :

Speed = Distance covered / Time taken

Calculation :

The time taken by the first person to complete a full lap is 1800 m / 9 m/s = 200 seconds.

​The time taken by the second person to complete a full lap is 1800 m / 6 m/s = 300 seconds.

​The LCM of 200 seconds and 300 seconds is 600 seconds, which means they will end up in the same initial positions after 600 seconds.

​Let the two persons cross each other for the first time after x seconds.

Then, total distance covered by 1st person = 9x and the 2nd person = 6x

Now, the total distance covered by both the person = Circumference of the track

⇒ 9x + 6x = 1800m ⇒ 15x = 1800m ⇒ x = 120 s

∴ The two persons will cross each other for the first time after 120 seconds.

The two persons will cross each other again after 120 seconds, and then again after 120 seconds, and so on.

So in 600 seconds, they will meet = 600/120 = 5 times

Therefore, the two persons will cross each other at 5 distinct points on the track.

Shortcut Trick

S1 = 9 m/s and S2 = 6 m/s 

S1/S2 = 3/2

They are running in opposite directions so they will meet each other at distinct points = 2 + 3 = 5 

A train leaves Kazipet at 5 a.m. and reaches Bangalore at 3 p.m. Another B train leaves Bangalore at 7 a.m. and reaches Kazipet at 5 p.m. When do the two trains meet? Assume that the trains travels at equal uniform speeds.

  1. 1 p.m.
  2. 12 noon
  3. 11 a.m.
  4. 10 a.m.

Answer (Detailed Solution Below)

Option 3 : 11 a.m.

Relative Speed Question 13 Detailed Solution

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Let speed be x km/hr

So, distance between stations = 10x km

When trains are travelling in opposite direction, relative speed = 2x km/hr.

Train A leaving at 5 am and train B at 7 am,

⇒ Distance travelled by A in 2 hrs = 2x

⇒ Remaining distance = 10x - 2x = 8x

⇒ Meeting time = 8x/2x = 4 hrs

∴ After 4 hrs, at 11 am train B will meet train A.

A policeman noticed a thief from 300 m. The thief started running and the policeman was chasing him. The thief and the policeman ran at the speeds of 8 km/h and 9 km/h, respectively. What was the distance between them after 3 minutes? 

  1. 225 m 
  2. 250 m  
  3. 300 m 
  4. 200  m

Answer (Detailed Solution Below)

Option 2 : 250 m  

Relative Speed Question 14 Detailed Solution

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Given:

Distance between policeman and thief in the starting = 300 m

Speed of policeman = 9 km/hr

Speed of thief = 8 km/hr

Concept: /Formula:

If the speed of a policeman and thief is x km/hr and y km/hr, then

Relative speed, if same directions = (x – y) km/hr

Distance between them after n hrs = (x – y) × n

1 km/hr = 5/18 m/sec

1 min = 60 sec

Calculation:

3 min = 3 × 60 = 180 seconds

Distance between policeman and thief in starting = 300 m

Relative speed of policeman and thief, if same directions = (9 – 8) = 1 × (5/18) = (5/18) m/sec

Distance covered in 180 seconds = (5/18) × 180 = 50 m

Distance between them after 180 seconds= 300 – 50 = 250 m

∴ Distance between them after 3 min is 250 m.

In a circular race of 5000 m, starting from the same point, the speeds of two contestants A and B are 36 km/h and 54 km/h, respectively. The difference between the time taken (in seconds) to meet for the first time when they are running in the opposite directions and when they are running in the same direction?

  1. 600
  2. 800
  3. 200
  4. 1000

Answer (Detailed Solution Below)

Option 2 : 800

Relative Speed Question 15 Detailed Solution

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Given

Circular race length: 5000 m

Speed of A: 36 km/h (or 10 m/s)

Speed of B: 54 km/h (or 15 m/s)

Concept:

Time = Distance/Speed. In opposite directions, the speeds add up; in the same direction, they subtract.

Solution:

Time taken when running in opposite directions = 5000/(10 + 15) ⇒ 200 seconds

Time taken when running in the same direction = 5000/(15 - 10) ⇒ 1000 seconds

The difference in times = 1000 - 200 ⇒ 800 seconds

Therefore, the difference between the times taken to meet for the first time in opposite and the same directions is 800 seconds.

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