Bunuel wrote:
Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?
(A) 6/7 hrs
(B) 12/7 hrs
(C) 4 hrs
(D) 6 hrs
(E) 12 hrs
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:What would we usually do in such a question? Two cars start from the same point and run in opposite directions – their speeds are given. This would remind us of relative speed. When two objects move in opposite directions, their relative speed is the sum of their speeds. So we might be tempted to do something like this:
Perimeter of the circle = 2\(\pi\)r = 2\(\pi\)*6 = 12? miles
Time taken to meet = Distance/Relative Speed = 12\(\pi\)/(6? + 8?) = 6/7 hrs
But take a step back and think – what does 6/7 hrs give us? It gives us the time taken by the two of them to complete one circle together. In this much time, they will meet somewhere on the circle but not at the starting point. So this is definitely not our answer.
The actual time taken to meet at point S will be given by 12\(\pi\)/(8\(\pi\) – 6\(\pi\)) = 6 hrs
This is what we mean by unexpected! The relative speed should be the sum of their speeds. Why did we divide the distance by the difference of their speeds? Here is why:
For the two objects to meet again at the starting point, obviously they both must be at the starting point. So the faster object must complete at least one full round more than the slower object. In every hour, car B – the one that runs at a speed of 8\(\pi\) mph covers 2\(\pi\) miles more compared with the distance covered by car A in that time (which runs at a speed of 6\(\pi\) mph). We want car B to complete one full circle more than car A. In how much time will car B cover 12\(\pi\) miles (a full circle) more than car A? In 12\(\pi\)/2\(\pi\) hrs = 6 hrs.