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

Given:

Rocket launch interval (T_launch) = 11.5 s

Time sound heard in train (T_heard) = 11 s

Speed of sound (V_s) = 330 m/s

Formula Used:

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

Calculations:

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

⇒ 11 × (330 + V_t) = 11.5 × 330

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

⇒ 330 + V_t = 30 × 11.5

⇒ 330 + V_t = 345

⇒ V_t = 345 - 330

⇒ V_t = 15 m/s

∴ The speed of the train is 15 m/s.

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.

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.

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

Given:

Speed of thief (v_thief) = 8 m/s

Speed of policeman (v_policeman) = 10 m/s

Initial separation (d) = 50 m

Formula used:

Time (t) = Distance / Relative Speed

Relative Speed = v_policeman - v_thief

Calculation:

Relative Speed = 10 - 8

⇒ Relative Speed = 2 m/s

Time (t) = 50 / 2

⇒ t = 25 s

∴ The correct answer is option (1).

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.

Concept used:

Speed × time = distance

Calculation:

In the 1^st 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

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

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.

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.

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

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 

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

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.

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.