Landon can climb a certain hill at a rate that is 2.5 mph slower than his rate coming down the hill. If it takes him 6 hours to climb the hill and 45 minutes to come down the hill, what is his rate coming down?
the distance is the same
Let s be his uphill speed
d/s = 6
d/(s+2.5) = .75
6s = .75(s+2.5)
6s = 3s/4 + 15/8
48s = 6s + 15
42s = 15
s = 15/42 = 5/14
So, he can go up at 5/14 mph and come down at 20/7 mph
6*5/14 = 30/14
3/4 * 20/7 = 60/28 = 30/14
Looks like the hill is 15/7 miles long.
To find Landon's rate coming down the hill, we can start by converting his time coming down the hill to hours.
Landon took 45 minutes to come down the hill, which is 45/60 = 0.75 hours.
Let's represent Landon's rate coming down the hill as x mph.
Given that Landon's rate climbing up the hill is 2.5 mph slower than his rate coming down, we can express his rate climbing up as x - 2.5 mph.
We know that time = distance / rate.
For Landon's climb up the hill, the time is 6 hours, the rate is x - 2.5 mph, and the distance is the same for both the climb and the descent.
Similarly, for Landon's descent down the hill, the time is 0.75 hours, the rate is x mph, and the distance is the same as the climb.
Since the distance is the same for both the climb and the descent, we can set up the equation:
Distance / Rate (up) = Time (up)
Distance / Rate (down) = Time (down)
The distances cancel out since they are the same, leaving us with:
1 / (x - 2.5) = 6 [Equation 1]
1 / x = 0.75 [Equation 2]
Now, let's solve the system of equations:
From Equation 1, we can rearrange it as (x - 2.5) = 1/6.
Simplifying, we get x = 1/6 + 2.5.
Combining like terms, x = 1/6 + 15/6.
Further simplifying, x = 16/6, which can be reduced to x = 8/3.
So Landon's rate coming down the hill is 8/3 mph.
Therefore, Landon's rate coming down the hill is approximately 2.67 mph.