Hearst Castle is 180 miles from Ventura by road. Two coaches leave Ventura at the same time, both heading for Hearst Castle, but one averages 5 mph faster than the other. If the faster coach reaches Hearst Castle half an hour earlier than the slower coach, what is the average speed of the faster coach?
To solve this problem, we can use the formula:
Distance = Speed * Time
Let's assume the speed of the slower coach is 'x' mph. This means the speed of the faster coach is 'x + 5' mph since it is 5 mph faster.
Now, let's calculate the time taken by both coaches to travel the distance of 180 miles:
For the slower coach:
Time = Distance / Speed
Time = 180 / x
For the faster coach:
Time = Distance / Speed
Time = 180 / (x + 5)
We are given that the faster coach reaches Hearst Castle half an hour earlier than the slower coach. So, the time taken by the slower coach plus half an hour (0.5) should be equal to the time taken by the faster coach:
180 / x + 0.5 = 180 / (x + 5)
Now, we need to solve this equation to find the value of 'x' (the speed of the slower coach).
To do that, we can start by cross-multiplying:
(180 / x) + 0.5 (180) = 180(x + 5)
Next, we distribute 0.5 to 180:
(180 / x) + 90 = 180(x + 5)
Now, let's multiply both sides of the equation by 'x' to remove the denominators:
180 + 90x = 180x + 900
Rearranging the equation:
90x - 180x = 900 - 180
-90x = 720
Dividing both sides by -90:
x = -720 / -90
x = 8
Therefore, the speed of the slower coach is 8 mph. Since the faster coach is 5 mph faster, its speed is:
8 + 5 = 13 mph
So, the average speed of the faster coach is 13 mph.