Two cars start from point A at the same time and drive around a circuit more than one mile in length. While they are driving laps around the circuit, each car must maintain a steady speed. One car is faster than the other and will pass the other car at certain points. The first pass occurs 150 yards from point A.

At what distance from A will one car pass the other again?

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300 yards from point A.

ReplyDeleteI just had to look at the math. So I got a little creative.

ReplyDeleteThe track is over a mile long, so lets call it exactly 1.5 miles. Hence, when car 1 passed car 2, car 1 had traveled 1.5 miles + 1.5 miles - 150 feet (2.97 miles). (But when you subtract the 1.5 miles car 1 did on his first lap, they would be at the same place) Car 2 had traveled 1.5 miles - 150 (1.47 miles). Since the time for both was the same, let's call it 3 min after the start. So we get a speed for car 1 of 59.43182 mph and a speed of car 2 of 29.43182 mph.

Now if you do some simple D = S * T formulas, you can find that at exactly 6 minutes, car 1 (faster car) had traveled 5.94318 miles. Car 2 (slower) had traveled 2.94318 miles. But since car 1 had traveled an additional 3 miles, while it was lapping car 2, they both reach 1.44318 miles around the track at exactly 6 minutes. 1.44318 / 1.5 * 5280 (feet in a mile) = 300 feet short of point A.

Yeah, whether it passes 150 feet

ReplyDeletein front ofpoint A or 150 feetbehindpoint A, the 2nd pass will occur 300 feet from point A.Good points all. 300 feet from point A is the answer.

ReplyDeleteUSAF Pilot, that was impressive work!

Can we visualise this problem as an analogue clock? I.e. every time the minute hand laps the hour hand, the point of overlap increases by a constant unit? So, if the first overlap on the racetrack is N, the second overlap point will be 2N?

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