Car A leaves city C at 5 pm and drives at a speed of 40 kmph. 2 hours later…

2024

Car A leaves city C at 5 pm and drives at a speed of 40 kmph. 2 hours later another car B leaves city C and drives in the same direction as car A. In how much time will car B be 9 km ahead of car A. Speed of car B is 60 kmph.

  1. A.

    4.25 hrs

  2. B.

    4.17 hrs

  3. C.

    4.30 hrs

  4. D.

    4.45 hrs

Attempted by 4 students.

Show answer & explanation

Correct answer: D

Concept: When two objects move in the same direction with a time gap between their starts, the speed at which the later one gains ground is the relative speed — the difference of their speeds. Any existing head-start distance, plus any extra lead required, is covered entirely at this relative speed.

Application:

  1. Car A starts at 5 pm at 40 kmph. By the time Car B starts, 2 hours later, Car A has already covered 40 × 2 = 80 km — this is the head-start gap Car B must first close.

  2. Car B's speed relative to Car A (same direction, so speeds subtract) is 60 − 40 = 20 kmph.

  3. To be 9 km AHEAD of Car A (not merely level with it), Car B must cover the 80 km head-start gap AND an extra 9 km lead — a total relative distance of 80 + 9 = 89 km.

  4. Time required, measured from when Car B starts = total relative distance ÷ relative speed = 89 ÷ 20 = 4.45 hours.

Cross-check:

  1. Let t = hours after Car B starts. Distance covered by Car A since it started = 40(t + 2), since Car A has been travelling 2 hours longer.

  2. Distance covered by Car B = 60t.

  3. Car B is 9 km ahead of Car A when 60t − 40(t + 2) = 9, i.e. 20t − 80 = 9, so 20t = 89 and t = 4.45 hours.

  4. Both methods agree at t = 4.45 hours, confirming the answer.

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