Suppose that a cyclist began a 450 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 7.5 hr and the car traveled 30.4 mph faster than the bicycle, find the average rate of each.
(v1 + v2)7.5 = 450
so
v1 + v2 = 60 miles/hour
and
v1 - v2 = 30.4
------------------ add
2 v1 = 90.4
v1 = 45.2 miles/hour
so
v2 = 14.8 miles/hour
To find the average rate of each, we can set up a system of equations based on the given information.
Let's denote the rate of the bicycle as "b mph" and the rate of the car as "c mph".
We know that the bicycle and car meet after 7.5 hours. This means that the total distance traveled by the bicycle and the car is equal to the total distance across the state, which is 450 miles.
The equation for the distance traveled by the bicycle is d = b * t, where d is the distance and t is the time. Similarly, the equation for the distance traveled by the car is d = c * t.
We are also given that the car traveled 30.4 mph faster than the bicycle. So we can write c = b + 30.4.
Since the total distance traveled is the same for both, we can set up the equation:
b * 7.5 = (b + 30.4) * 7.5
Simplifying this equation, we get:
7.5b = 7.5b + 228
Subtracting 7.5b from both sides, we find:
0 = 228
This is not possible, which means there is no solution to this equation.
Therefore, there is no average rate for the bicycle and car that satisfies the given conditions.