Question

A motorcyclist and a cyclist travel from a to b. they leave simultaneously from a at 12 noon. The motor cyclist arrives at b after 1 1/2 hours and the cyclist arrives 30 mins later. At what time was the cyclist twice as far from b as the motor cyclist was.

Answer 1

The time, when the cyclist was twice as far from b as the motorcyclist was is 1.12 p.m.

Explanation:

The time both the motorcyclist and the cyclist leave from a = 12 noon

The time it takes the motor cyclist to arrive at b = 1 1/2 hours = 1 hour, 30 minutes

The time it takes the cyclist to arrive at b = 30 min later than the motorcyclist = 30 minutes + 1 hour 30 minutes = 2 hours

The speed of the motorcyclist = x/1.5 mph

The speed of the cyclist = x/2 mph

Let the time, in hours, when the cyclist was twice as far from b as the motor cyclist was be t, and let c represent the location of the cyclist at time, t, we have;

The ratio of the speed of the two cyclist is 3:4

t × b/2 = b - (b - c) and t × b/1.5 = b - (b - c)/2

c/(t/2) = (b/2 + c/2)/(t/1.5)

2·c = 1.5·b/2 + 1.5·c/2

2·c - 1.5·c/2 = 1.5·b/2

1.25·c = 0.75·b

c = 0.6·b

Substituting gives;

t × b/2 = b - (b - 0.6·b)

t = 0.6·b ×2/b = 1.2

Therefore, the time, in hours, when the cyclist was twice as far from b as the motorcyclist was = t = 1.2 hours from the starting time or 1.12 p.m.