Two cyclists left simultaneously from cities A and B heading towards each other at constant rates and met in 5 hours. The rate of the cyclist from A was 3 mph less than the rate of the other cyclist. If the cyclist from B had started moving 30 minutes later than the other cyclist, then the two cyclists would have met 31.8 miles away from A. What is the distance between A and B, in miles?
Let d = distance from A to B
If a is A's speed, then a+3 = B's speed
d = 5a + 3(a+3)
since distance = speed * time, then at time t hours from when A started out
at = 31.8
(a+3)(t-1/2) = d-31.8
Solve those three equations to find d
To solve this problem, we need to break it down into steps and use the given information to find the distance between cities A and B.
Let's start by assigning variables to the unknowns in the problem:
Let x be the rate of the cyclist from B (in mph).
The rate of the cyclist from A would be x - 3 mph (since the rate from A is 3 mph less than B's rate).
Now we can calculate the distance each cyclist traveled:
The distance traveled by the cyclist from A would be D1 = (x - 3) mph * 5 hours.
The distance traveled by the cyclist from B would be D2 = x mph * 5 hours.
Since the two cyclists met after 5 hours, their total distance traveled must be the same (D1 = D2).
Therefore, we have the equation: (x - 3) mph * 5 hours = x mph * 5 hours.
Simplifying this equation, we get: 5(x - 3) = 5x.
Expanding the equation: 5x - 15 = 5x.
Subtracting 5x from both sides: -15 = 0.
Uh-oh! We have a contradiction, which means there is no solution to this equation and the problem as stated is impossible.
This contradiction suggests there might be an error in the problem statement. Please check the given information again or provide any additional details if available.