Problema Solution
Two cyclists, 192 mi apart, start riding towards each other at the same time. One cycles twice as fast as the other. If they meet 4 hr later, at what average speed is each cyclist traveling?
Answer provided by our tutors
distance = speed * time
time = distance/speed
let 'x' represent the distance covered by the slow cyclist, then '192-x' represents the distance covered by the fast cyclist
let 'v' represent the speed of the slower cyclist, then '2v' represents the speed of the faster cyclist
the cycle the same amount of time, so we set their distance/speed expressions equal:
x/v = (192-x)/2v
solving for 'x' we have x=64
the slow cyclist travelled 64 miles
64 = s*4
s = 64/4 = 16 mph
the fast cyclist travelled 192-64 = 128 miles
128/4 = 32 mph
the slow cyclist travelled at 16mph, the fast cyclist travelled at 32mph