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

Given:

Train X overtakes Train Y in 80 sec

Train Y passes a man in 14 sec

Length of Train X = 21l, Length of Train Y = 19l

Speed of Train X = 3v, Speed of Train Y = 2v

Formula used:

Overtaking Time = (Length of X + Length of Y) ÷ (Speed of X − Speed of Y)

Speed = Distance ÷ Time

Time to cross bridge = (Length of Train + Length of Bridge) ÷ Speed of Train

Calculations:

Overtaking time:

⇒ (21l + 19l) ÷ v = 80

⇒ 40l ÷ v = 80

⇒ l ÷ v = 2 ⇒ l = 2v

⇒ v = a and b = 21l - 19l = 2l

Bridge length = 120b = 120 × 2l = 240l

Length of Train X = 21l

Total distance = 240l + 21l = 261l

Speed of Train X = 3v = 3a = 3l/2

⇒ Time = 261l ÷ (3l/2) = 261 × 2 ÷ 3 = 174

∴ Time taken by Train X to cross the bridge = 174 seconds.

Given:

Speed of both cabs = 54 km/hr

Time gap between cab departures = 29 minutes

Man meets the cabs at an interval of 12 minutes

Formula used:

Distance between cabs = Relative speed (man & cab) × time between meetings

Calculations:

⇒ Distance between cabs = 54 × (29 ÷ 60) = 54 × 0.4833 = 26.1 km

⇒ Relative speed when man meets cab = Distance ÷ (12 ÷ 60) = 26.1 ÷ 0.2 = 130.5 km/hr

⇒ Man’s speed = Relative speed − Cab speed = 130.5 − 54 = 76.5 km/hr

∴ The speed of the man is 76.5 km/hr.

Given:

Speed of cabs (V_c) = 74 km/hr

Time interval between cabs starting (T_c) = 28 minutes

Time interval at which the man meets the cabs (T_m) = 10 minutes

Formula Used:

Distance = Speed × Time

Relative speed when moving in opposite directions = Sum of individual speeds

Calculation:

T_c = 28 minutes = $\frac{28}{60}$ hours = $\frac{7}{15}$ hours

T_m = 10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours

The distance between two consecutive cabs is the distance covered by the first cab in the interval before the second cab starts.

Distance between cabs = V_c × T_c

Distance between cabs = $74\times \frac{28}{60}$ km

Let the speed of the man be V_m km/hr.

Since the man is moving in the opposite direction to the cabs, their relative speed is the sum of their individual speeds (V_c + V_m).

The man meets the two cabs at an interval of T_m (10 minutes). This means the distance between the cabs is covered by their relative speed in 10 minutes.

Distance between cabs = (V_c + V_m) × T_m

Distance between cabs = $(74+V_m)\times \frac{10}{60}$ km

Equate the expressions for the distance between cabs and solve for V_m.

$74\times \frac{28}{60}=(74+V_m)\times \frac{10}{60}$

74 × 28 = (74 + V_m) × 10

2072 = 740 + 10V_m

2072 - 740 = 10V_m

1332 = 10V_m

$V_m=\frac{1332}{10}$

V_m = 133.2 km/hr

∴ The speed of the man is 133.2 km/hr.

Given:

Distance = 1 km

Initial speed = X km/h

Speed reduced by = Y km/h

Time taken = Z hours

Formula used:

Time = Distance / Speed

Calculation:

The man's initial speed is X km/h, and it is reduced by Y km/h due to sticky ground.

Actual Speed = Initial Speed - Speed Reduced

⇒ Actual Speed = (X - Y) km/h

Z = 1 / (X - Y)

Rearranging the equation: Z × (X - Y) = 1

⇒ ZX - ZY = 1

Using the formula Z × (X - Y) = 1:

⇒ ZX - ZY = 1

∴ The correct answer is option (3).

Calculation

Car A speed = 72 km/h

Car B speed = 54 km/h

They are moving in opposite directions

Distance between them = 693 km

They meet after t hours

When two objects move in opposite directions, their relative speed is the sum of their speeds:

Relative Speed = 72 + 54 = 126 km/h

Time = [Distance/Relative Speed] = 693/126 = 5.5 hours

So, t + 2.5 = 5.5 + 2.5 = 8 

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 = ${\displaystyle \frac{distance}{speed}}$

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 

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.

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.