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# In Constantly Passing

##### Age 14 to 16Challenge 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.