Four vehicles travelled on a road with constant velocities. The car
overtook the scooter at 12 o'clock, then met the bike at 14.00 and
the motorcycle at 16.00. The motorcycle met the scooter at 17.00
then it overtook the bike at 18.00. At what time did the bike and
the scooter meet?
Brian swims at twice the speed that a river is flowing, downstream
from one moored boat to another and back again, taking 12 minutes
altogether. How long would it have taken him in still water?
At Holborn underground station there is a very long escalator. Two
people are in a hurry and so climb the escalator as it is moving
upwards, thus adding their speed to that of the moving steps. ...
How many steps are there on the escalator?
Thank Justin from Skyview High School, Billings, MT,
USA for this solution and well
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
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,
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.