Problema Solution
Suppose that a cyclist began a 429 mi 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 6.5 hr and the car traveled 33.6 mph faster than the bicycle, find the average rate of each.
Answer provided by our tutors
Let
d = 429 mi the distance
t = 6.5 hr the time of the travel
v1 = the average rate of the cyclist
v2 = the average rate of the car
The car traveled 33.6 mph faster than the bicycle:
v2 = 33.6 + v1
Since average rate = distance/time => distance = time*avg.rate:
v1t + v2t = d
6.5v1 + 6.5v2 = 429
We have the following system of equations:
v2 = 33.6 + v1
6.5v1 + 6.5v2 = 429
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click here to see the system of equations solved for v1 and v2
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v1 = 16.2 mph
v2 = 49.8 mph
The average rate of the bicycle is 16.2 mph.
The average rate of the car is 49.8 mph.