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A thief is spotted by a policeman from a distance of 210 m. When the policeman starts the chase, the thief also starts running. If the speed of the thief is 25 km/h and that of the policeman is 32 km/h, then how far would the thief have run (in m) before he is overtaken?
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750
- Relative Speed: The relative speed of the policeman catching up with the thief is the difference between their speeds. It can be calculated as 32 km/h - 25 km/h = 7 km/h.
- Convert to meters per second (m/s): To calculate the time, we need to convert the relative speed from km/h to m/s. $$7 \text{ km/h} = \frac{7 \times 1000}{60 \times 60} \text{ m/s} = \frac{35}{18} \text{ m/s}$$.
- Time to Close Distance: The distance that needs to be closed is 210 meters. The time taken can be calculated using the formula $$ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{210}{\frac{35}{18}}$$.
- Solve for Time: The calculation gives time \( = \frac{210 \times 18}{35} \approx 108 \text{ seconds}\).
- Distance Run by Thief: The thief's speed is 25 km/h \(= \frac{25 \times 1000}{60 \times 60} = \frac{125}{18} \text{ m/s}\). The distance run by the thief is \( = \text{Speed} \times \text{Time} = \frac{125}{18} \times 108 \approx 750 \text{ meters}\).
- Conclusion: Based on calculations, the thief would have run 750 meters before being overtaken.
- Correct Answer: 750 meters.
By: Parvesh Mehta ProfileResourcesReport error
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