Answer:
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.