Speed, Time and Distance
- Two men set out the same time to walk towards each other from two points A and B, 72 km apart. The first man walks at the rate of 4 km per hr. The second man walks 2 km in the first hour,
Find the time after which the two men will meet.2 1 km in the second hour, 3 km in the third hour and so on. 2
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Let the two men meet after t hours.
Distance covered by the first man starting from A = 4 t km.
Distance covered by the second man starting from B= 2 + 2.5 + 3 +......+ 
2 + 
t − 1 
2
This is an arithmetic series of t terms with 1/2 as common difference.
∴ By applying formulaS = n 2a + (n − 1)d 2
Where, n = no. of terms
a = first term
d = common difference
We have its sumS = t (2 × 2) + (t − 1) × 1 2 2 = 2t + t2 − t 4
Total distance covered by two men == 4t + 2t + t2 − t = 72 4
or= 6t + t2 − t = 72 4
or 24t + t2 – t = 288
or t2 + 23t – 288 = 0
or t2 – 9t + 32t – 288 = 0
or t (t–9) + 32 (t – 9) = 0
or (t – 9) (t + 32) = 0
∴ Either t – 9 = 0 ⇒ t = 9
or (t + 32) = 0 ⇒ t = – 32
Time cannot be negative.
Hence, the two men will meet after 9 hrs.Correct Option: B
Let the two men meet after t hours.
Distance covered by the first man starting from A = 4 t km.
Distance covered by the second man starting from B= 2 + 2.5 + 3 +......+ 2 + t − 1 2
This is an arithmetic series of t terms with 1/2 as common difference.
∴ By applying formulaS = n 2a + (n − 1)d 2
Where, n = no. of terms
a = first term
d = common difference
We have its sumS = t (2 × 2) + (t − 1) × 1 2 2 = 2t + t2 − t 4
Total distance covered by two men == 4t + 2t + t2 − t = 72 4
or= 6t + t2 − t = 72 4
or 24t + t2 – t = 288
or t2 + 23t – 288 = 0
or t2 – 9t + 32t – 288 = 0
or t (t–9) + 32 (t – 9) = 0
or (t – 9) (t + 32) = 0
∴ Either t – 9 = 0 ⇒ t = 9
or (t + 32) = 0 ⇒ t = – 32
Time cannot be negative.
Hence, the two men will meet after 9 hrs.
- A man is standing on a railway bridge which is 50 metres long. He finds that a train crosses the bridge
Find the length of the train and its speed.in 4 1 seconds but himself in 2 seconds. 2
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Let the length of the train be x metres
Then, the time taken by the train to cover (x + 50) metres is4 1 seconds 2 = x + 50 9 2 or 2x + 100 m per second ...(i) 9
Again, the time taken by the train to cover x metres in 2 seconds.∴ Speed of the train = x metre per second ...(ii) 2
From equations (i) and (ii), we have2x + 100 = x 9 2
⇒ 4x + 200 = 9x
⇒ 5x = 200
⇒ x = 40
∴ Length of the train = 40 metres
∴ Speed of the train= x = 40 = 20 m per sec. 2 2 Correct Option: C
Let the length of the train be x metres
Then, the time taken by the train to cover (x + 50) metres is4 1 seconds 2 = x + 50 9 2 or 2x + 100 m per second ...(i) 9
Again, the time taken by the train to cover x metres in 2 seconds.∴ Speed of the train = x metre per second ...(ii) 2
From equations (i) and (ii), we have2x + 100 = x 9 2
⇒ 4x + 200 = 9x
⇒ 5x = 200
⇒ x = 40
∴ Length of the train = 40 metres
∴ Speed of the train= x = 40 = 20 m per sec. 2 2
- Two places A and B are 162 kms apart. A train leaves A for B and at the same time another train leaves B for A. The two trains meet at the end of 6 hours. If the train travelling from A to B travels 8 km per hr. faster than the other, find the speed of the faster train.
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Both trains meet after 6 hours.
∴ The relative speed of two trains= 162 = 27 km per hr. 6
The speed of the slower train starting from B= 27 − 8 = 19 =9 1 km per hr. 2 2 2
∴ The speed of the faster train= 9 1 + 8 = 17 1 km per hr. 2 2 Correct Option: D
Both trains meet after 6 hours.
∴ The relative speed of two trains= 162 = 27 km per hr. 6
The speed of the slower train starting from B= 27 − 8 = 19 =9 1 km per hr. 2 2 2
∴ The speed of the faster train= 9 1 + 8 = 17 1 km per hr. 2 2
- A train running at 25 km per hour take 18 seconds to pass a platform. Next, it takes 12 seconds to pass a man walking at the rate of 5 km per hr. in the same direction. Find the length of the platform.
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Let the length of train be x metres and the length of platform be y metres.
Speed of the train = 25 × 5 m/sec 18 = 125 m per sec. 18
Time taken by train to pass the platform= (x + y) × 18 sec. 125 ∴ (x + y) × 18 = 18 125
or, x + y = 125 ...(i)
Speed of train relative to man = (25 + 5) km per hr.= 30 × 5 m per sec. 18 = 25 m per sec. 3
Time taken by the train to pass the man= x × 3 sec. 25 = 3x sec. 25 ∴ 3x = 12 25 ⇒ x = 25 × 12 = 100 metres 3
Putting x = 100 in equation (i),
we get, y = 25 metres.
∴ Length of train = 100 metres
and length of the platform = 25 metres.Correct Option: A
Let the length of train be x metres and the length of platform be y metres.
Speed of the train = 25 × 5 m/sec 18 = 125 m per sec. 18
Time taken by train to pass the platform= (x + y) × 18 sec. 125 ∴ (x + y) × 18 = 18 125
or, x + y = 125 ...(i)
Speed of train relative to man = (25 + 5) km per hr.= 30 × 5 m per sec. 18 = 25 m per sec. 3
Time taken by the train to pass the man= x × 3 sec. 25 = 3x sec. 25 ∴ 3x = 12 25 ⇒ x = 25 × 12 = 100 metres 3
Putting x = 100 in equation (i),
we get, y = 25 metres.
∴ Length of train = 100 metres
and length of the platform = 25 metres.
- Two trains 200 metres and 175 metres long are running on parallel lines.
They take 7 1 seconds.when running in opposite directions 2
to pass each other. Find their speeds in km per hour.and 37 1 seconds when running in the same direction 2
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Let the speed of the train be x metre per sec. and y metre per sec. respectively.
Sum of the length of the trains = 200 + 175 = 375 metres
Case : I
When the trains are moving in opposite directions
Relative speed = (x + y) m per sec.
In this case the time taken by the trains to cross each other= 375 sec. x + y ∴ 375 = 15 x + y 2
⇒ x + y = 50 ...(i)
Case : II
When the trains are moving in the same direction.
Relative speed = (x – y) m per sec.
In this case, the time taken by the trains to cross each other= 375 sec. x − y ∴ 375 = 75 x − y 2
⇒ x – y = 10 ...(ii)
Now, x + y = 50x – y = 10 _________ 2x = 60
⇒ x = 30
Putting this value in equation (i),
we have
y = 50 – 30 = 20
∴ Speed of trains = 30 m per sec.= 30 × 18 = 108 km per hr. 5 and 20 m per sec. = 20 × 18 = 72 km per hr. 5 Correct Option: B
Let the speed of the train be x metre per sec. and y metre per sec. respectively.
Sum of the length of the trains = 200 + 175 = 375 metres
Case : I
When the trains are moving in opposite directions
Relative speed = (x + y) m per sec.
In this case the time taken by the trains to cross each other= 375 sec. x + y ∴ 375 = 15 x + y 2
⇒ x + y = 50 ...(i)
Case : II
When the trains are moving in the same direction.
Relative speed = (x – y) m per sec.
In this case, the time taken by the trains to cross each other= 375 sec. x − y ∴ 375 = 75 x − y 2
⇒ x – y = 10 ...(ii)
Now, x + y = 50x – y = 10 _________ 2x = 60
⇒ x = 30
Putting this value in equation (i),
we have
y = 50 – 30 = 20
∴ Speed of trains = 30 m per sec.= 30 × 18 = 108 km per hr. 5 and 20 m per sec. = 20 × 18 = 72 km per hr. 5
