average rate

Question

Suppose that a cyclist began a 627mi ride across a state at the western edge of the state, at the same time that a car traveling toward it leaves the eastern end of the state. If the bicycle and car met after 9.5hr and the car traveled 31.6mph faster than the bicycle, find the average rate of each.

The car's average rate is mph?
The bicycle's average rate is mph?

Answer 1

v = speed of bicycle in mph

u = speed of car in mph = v + 31.6

t = time they meet = 9.5 h

 

Suppose they meet at a distance x from the west end.  It is the same x for both, so

 

x = vt = v(9.5) =9.5v

x = 627 - ut = 627 - (v + 31.6)t = 627 - (v + 31.6)(9.5) = 326.8 - 9.5v

9.5v = 326.8 - 9.5v

v = 17.2 mph

u = v + 31.6 = 17.2 + 31.6 = 48.8 mph

 

Speed of bicycle:  17.2 mph

Speed of of car:  48.8 mph