Introduction / Context:
Chase problems are a standard application of relative speed. In this situation, a thief has a head start, and a policeman runs faster from behind. We know the initial gap and both speeds. The question asks how far the thief will have run by the time he is caught. This requires computing the relative speed between them and then using that to find the time to close the gap, and finally the distance the thief covers during that time.
Given Data / Assumptions:
- Initial distance between thief and policeman = 150 m.
- Speed of thief = 7 km/h.
- Speed of policeman = 9 km/h.
- Both start moving at the same instant and in the same direction.
- Speeds remain constant until the thief is caught.
Concept / Approach:
When two objects move in the same direction, the closing speed (relative speed) is the difference between their speeds. Once we know the relative speed, we can compute the time needed to cover the initial gap and then the total distance run by the thief. The steps are:
relative speed = speed of policeman - speed of thief
time to catch = initial gap / relative speed
distance run by thief = speed of thief * time to catch
Step-by-Step Solution:
Step 1: Compute the relative speed.
Speeds are given in km/h. Relative speed = 9 - 7 = 2 km/h.
Step 2: Convert the gap to kilometres.
Initial gap = 150 m = 150 / 1000 km = 0.15 km.
Step 3: Compute the time to catch up.
time (hours) = gap / relative speed = 0.15 / 2 = 0.075 hours.
Step 4: Compute the distance run by the thief.
Thief's speed = 7 km/h.
distance run by thief = 7 * 0.075 km.
7 * 0.075 = 0.525 km.
Convert to metres: 0.525 km = 0.525 * 1000 = 525 m.
Therefore, the thief will have run 525 m by the time he is caught.
Verification / Alternative check:
We can also compute the distance covered by the policeman in this time to confirm. In 0.075 hours, at 9 km/h, the policeman travels:
distance = 9 * 0.075 = 0.675 km = 675 m.
The difference between their distances should be the initial gap of 150 m:
675 m - 525 m = 150 m.
This matches the initial separation, confirming our computations.
Why Other Options Are Wrong:
420 m corresponds to a smaller time, implying a higher relative speed than is actually the case.
630 m would give a larger difference between their distances than 150 m at the moment of capture.
315 m is too small and does not satisfy the closing gap condition with the given speeds.
450 m also fails to maintain the required 150 m difference at the start and zero gap at the capture moment.
Only 525 m fits all the conditions of the chase.
Common Pitfalls:
Some learners mistakenly add the speeds when the objects move in the same direction, instead of using the difference. Another common error is to forget to convert metres to kilometres or vice versa, leading to inconsistent units. Additionally, a few students directly compute the distance covered by the policeman and misinterpret that as the thief's distance. Always remember that we are asked for the distance run by the thief, not the policeman.
Final Answer:
The thief will have run
525 m before he is caught by the policeman.
Discussion & Comments