A thief is spotted by a policeman from a distance of 100 m. When the policeman starts the chase, the thief also starts chasing. If the speed of the thief is 8 km/hr and that of the policeman is 10 km/hr, then how far will the thief have to run before he is overtaken?
400 m
Understanding the Thief and Policeman Chase Problem
This problem involves two moving objects, a thief and a policeman, where the policeman is chasing the thief. We are given their speeds and the initial distance between them. We need to find out how far the thief runs before the policeman catches him. This is a classic example of a relative speed problem.
Calculating Relative Speed in the Chase
When one object chases another in the same direction, the speed at which the distance between them changes is called their relative speed. The relative speed is the difference between the speeds of the faster object (policeman) and the slower object (thief).
First, let's list the given information:
- Initial distance between thief and policeman = 100 m
- Speed of the thief = 8 km/hr
- Speed of the policeman = 10 km/hr
To work with consistent units (meters and seconds), it's best to convert the speeds from km/hr to m/s. We know that 1 km = 1000 m and 1 hour = 3600 seconds. So, 1 km/hr = \(\frac{1000}{3600}\) m/s = \(\frac{5}{18}\) m/s.
- Thief's speed in m/s: \(8 \times \frac{5}{18} = \frac{40}{18} = \frac{20}{9}\) m/s
- Policeman's speed in m/s: \(10 \times \frac{5}{18} = \frac{50}{18} = \frac{25}{9}\) m/s
Now, let's calculate the relative speed of the policeman with respect to the thief:
Relative Speed = Speed of Policeman - Speed of Thief
Relative Speed = \(\frac{25}{9}\) m/s - \(\frac{20}{9}\) m/s = \(\frac{25-20}{9}\) m/s = \(\frac{5}{9}\) m/s
This relative speed is the speed at which the policeman closes the initial gap of 100 m.
Determining the Time to Overtake
The policeman needs to cover the initial distance of 100 m at the relative speed to catch the thief. We can use the formula: Time = Distance / Speed.
Time taken to overtake = \(\frac{\text{Initial Distance}}{\text{Relative Speed}}\)
Time = \(\frac{100 \text{ m}}{\frac{5}{9} \text{ m/s}}\)
Time = \(100 \times \frac{9}{5}\) seconds
Time = \(20 \times 9\) seconds
Time = 180 seconds
So, the policeman will catch the thief after 180 seconds.
Calculating the Distance Run by the Thief
In the 180 seconds that the chase lasts, the thief continues running at his speed of 8 km/hr (or \(\frac{20}{9}\) m/s). We can find the distance the thief runs using the formula: Distance = Speed × Time.
Distance run by thief = Thief's Speed × Time taken to overtake
Distance run by thief = \(\frac{20}{9}\) m/s \(\times\) 180 seconds
Distance run by thief = \(\frac{20 \times 180}{9}\) meters
Distance run by thief = \(20 \times 20\) meters (since 180/9 = 20)
Distance run by thief = 400 meters
Thus, the thief will have run 400 m before the policeman overtakes him.
Verification
Let's verify the distance run by the policeman in the same time (180 seconds) at his speed of \(\frac{25}{9}\) m/s:
Distance run by policeman = Policeman's Speed × Time taken to overtake
Distance run by policeman = \(\frac{25}{9}\) m/s \(\times\) 180 seconds
Distance run by policeman = \(\frac{25 \times 180}{9}\) meters
Distance run by policeman = \(25 \times 20\) meters
Distance run by policeman = 500 meters
The policeman started 100 m behind the thief. In 180 seconds, the thief ran 400 m. The policeman ran 500 m. The difference in distance covered is 500 m - 400 m = 100 m, which is the initial distance between them. This confirms our calculation is correct.
Revision Table: Thief and Policeman Problem
Here's a quick summary of the steps and calculations:
| Concept | Formula/Value | Calculation |
|---|---|---|
| Initial Distance | \(D_{initial}\) | 100 m |
| Thief's Speed | \(V_{thief}\) | 8 km/hr = \(\frac{20}{9}\) m/s |
| Policeman's Speed | \(V_{police}\) | 10 km/hr = \(\frac{25}{9}\) m/s |
| Relative Speed | \(V_{relative} = V_{police} - V_{thief}\) | \(\frac{25}{9} - \frac{20}{9} = \frac{5}{9}\) m/s |
| Time to Overtake | \(T = \frac{D_{initial}}{V_{relative}}\) | \(\frac{100}{\frac{5}{9}} = 100 \times \frac{9}{5} = 180\) seconds |
| Distance run by Thief | \(D_{thief} = V_{thief} \times T\) | \(\frac{20}{9} \times 180 = 400\) meters |
Additional Information: Relative Speed Concepts
Understanding relative speed is crucial for solving problems involving objects moving towards or away from each other, or chasing each other.
- Objects moving in the same direction: The relative speed is the difference between their individual speeds (\(|V_1 - V_2|\)). This is the rate at which the distance between them changes.
- Objects moving in opposite directions (towards each other): The relative speed is the sum of their individual speeds (\(V_1 + V_2\)). This is the rate at which the distance between them decreases.
- Objects moving in opposite directions (away from each other): The relative speed is the sum of their individual speeds (\(V_1 + V_2\)). This is the rate at which the distance between them increases.
Always ensure that the units of distance and speed are consistent before performing calculations.
Similar Questions
Two men, A and B run a 4 km race on a course 0.25 km round. If their speeds are in the ratio 5 : 4, how often does the winner pass the other?
Three cars A, B and C started from a point at 5 p.m., 6 p.m. and 7 p.m. respectively and travelled at uniform speeds of 60 km/hr., 80 km/hr. and x km/hr. respectively in the same direction. If all the three met at another point at the same instant during their journey, then what is the value of x?
X, Y and Z travel from the same place with uniform speeds 4 km/hr, 5 km/hr and 6 km/hr respectively. Y starts 2 hours after X. How long after Y must Z start in order that they overtake X at the same instant?
A car did a journey in t hours. Had the average speed been x kmph greater, the journey would have taken y hours less. How long was the journey?
In covering certain distance, the average speeds of X and Y are in the ratio 4 : 5. If X takes 45 minutes more than Y to reach the destination, then what is the time taken by Y to reach the destination?
In a race of 1000 m, A beats B by 150 m, while in another race of 3000 m, C beats D by 400 m. Speed of B is equal to that of D. (Assume that A, B, C and D run with uniform speed in all the events). If A and C participate in a race of 6000 m, then which one of the following is correct?
When the speed of a train is increased by 20%, it takes 20 minutes less to cover the same distance. What is the time taken to cover the same distance with the original speed?
The speeds of three cars are in the ratio 2 : 3 : 4. What is the ratio between the time taken by these cars to travel the same distance?
A bike consumes 20 mL of petrol per kilometre, if it is driven at a speed in the range of 25 – 50 km/hour and consumes 40 mL of petrol per kilometre at any other speed. How much petrol is consumed by the bike in travelling a distance of 50 km, if the bike is driven at a speed of 40 km/hour for the first 10 km, at a speed of 60 km/hour for the next 30 km and at a speed of 30 km/hour for the last 10 km?
In a race of 1000 m, A beats B by 100 m or 10 seconds. If they start a race of 1000 m simultaneously from the same point and if B gets injured after running 50 m less than half the race length and due to which his speed gets halved, then by how much time will A beat B?
Important Questions from Relative Speed
In a circular race of 2500 m, a man and a woman start from a point towards opposite directions with speeds of 37 km/h and 35 km/h, respectively. After how much time from the start of the race will they meet for the first time?
- X and Yrun a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3 : 2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
A train of length 600 meters passes another train of length 1000 meters moving in the opposite direction in 96 seconds. If the speed of both the trains is the same, then what is the speed of each train?
The distance between Delhi and Patna is $480$ km. A train starts from Delhi and travels towards Patna at a speed of $60$ kmph. Another train starts from Patna towards Delhi at a speed of $80$ kmph, but starts $1$ hour after the first train. In how much time, after the first train started, will they meet each other?
Ram traveling at the speed of 3 km/h reaches 15 minutes late. Had he walked at 4 km/h, he would have reached 15 minutes earlier. How much distance does Ram have to cover?
