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In Constantly Passing

Age 14 to 16
Challenge Level

Thank Justin from Skyview High School, Billings, MT, USA for this solution and well done!

Rate, time and distance are connected by the equation r =d/t .

Call the rate (or speed) of the car r c and the rate of every bus r b . Each bus is a constant distance from the bus preceding it and the bus following it; call this distance d.

For a bus approaching on the other highway and coming towards the car, the rate of the bus relative to the car, considering the bus still, is (r b + r c ). This rate multiplied by the time it takes (three minutes) for the car to close the gap between it and the bus is equal to d, hence:

3(r b + r c ) = d.

The rate of the bus which is travelling in the same direction as the car, relative to the rate of the car (considering the car still) is (r b - r c ). This rate multiplied by the time it takes (six minutes) for the bus to close the gap between it and the car is also equal to d 1 , hence

6(r b - r c ) = d.

Multiplying the first equation by 2 and add the two equations, one obtains

12r b = 3d.

But the distance d between the buses divided by the rate of the bus is equal to the time interval between the buses therefore the time interval = d/r b = 4 minutes.

So the buses leave the depot at intervals of 4 minutes.