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.
Answer: D. 4.45 hrs — 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…
- A.
4.25 hrs
- B.
4.17 hrs
- C.
4.30 hrs
- D.
4.45 hrs
Attempted by 28 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:
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.
Car B's speed relative to Car A (same direction, so speeds subtract) is 60 − 40 = 20 kmph.
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.
Time required, measured from when Car B starts = total relative distance ÷ relative speed = 89 ÷ 20 = 4.45 hours.
Cross-check:
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.
Distance covered by Car B = 60t.
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.
Both methods agree at t = 4.45 hours, confirming the answer.