A thief is noticed by a policeman from a distance of 92 m. The thief starts running and the policeman chases him. The thief and the policeman run at the speed of 90 km/h and 104.4 km/h, respectively. In how much time will the policeman catch the thief?

Correct — D. Both men run the same way, so the gap closes at the difference of the speeds, not their sum.

Relative speed = 104.4 − 90 = 14.4 km/h
In m/s: 14.4 × 5⁄18 = 4 m/s

Time = gap ÷ closing speed = 92 ÷ 4 = 23 seconds → option (d)

Why the others are wrong

• (a)
  21 seconds — 21 seconds would need a closing speed of 92⁄21 ≈ 4.38 m/s. The two given speeds differ by 14.4 km/h, which is exactly 4 m/s.
• (b)
  26 seconds — 26 seconds at 4 m/s covers 104 m, more than the gap. The policeman has only 92 m to make up.
• (c)
  22 seconds — 22 seconds closes 88 m — four metres short. The catch happens the instant the full 92 m is covered, at 23 s.

This is relative speed in one dimension. When two bodies move in the same direction the gap between them shrinks at the difference of their speeds; when they move towards each other it shrinks at the sum.

Here the closing speed is 14.4 km/h, and the head start is quoted in metres, so the speed has to be converted before dividing.

Multiply km/h by 5⁄18 to get m/s. The numbers in this item are chosen so that 14.4 km/h lands on a clean 4 m/s.

Key facts

• Same-direction chase: closing speed is the difference of the speeds, 14.4 km/h here.
• 1 km/h = 5⁄18 m/s, so 14.4 km/h = 4 m/s exactly.
• The 92 m head start is a distance, so both quantities must be in metres and seconds before dividing.

Common traps

• Adding the speeds instead of subtracting, which gives a closing speed of 194.4 km/h
• Dividing 92 by 14.4 without converting to m/s
• Answering the catch-up time when the item asks for the gap after a stated interval

No directly related past PYQ was found.