SOLUTION: HELP!!! I have worked this problem and can't get it. Problem: Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and

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Question 37145: HELP!!! I have worked this problem and can't get it.
Problem:
Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and in the same amount of time runs 4 miles against the wind. What is the rate of the wind?
Found 3 solutions by fractalier, mbarugel, AnlytcPhil:
Answer by fractalier(6550)   (Show Source):
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Okay. The governing equation for this kind of problem is RT = D. Rate times time equals distance. His rate (with no wind) is given as 5 mph. Then we set up two equations, one with the wind and one against:
(R + W)T = D
(R - W)T = D
now substitute in what you know
(5 + W)T = 10
(5 - W)T = 4
If we solve the second one for T
T = 4 / (5 - W)
and then plug it in to the first equation, we get
(5 + W)( 4 / (5 - W)) = 10
Solving this we get W = 15/7 or about 2.143 mph, the wind speed.

Answer by mbarugel(146)   (Show Source):
You can put this solution on YOUR website!
Hello!
We must use the following relationship in order to answer this:
If an object is moving at a speed of V miles per hour, then the time (in hours) it takes to travel a distance D (in miles) is D/V.
Now, we know that Jim can run at 5 mph without wind. Let's call X to the speed of the wind. So Jim's speed is 5 + X mph when he goes with the wind, and 5 - X mph when he goes against it.
Using the aforementioned relationship, and the fact that he takes the same time to make 10 miles with the wind and 4 miles against the wind, we get the following equation:

So we just isolate X:

So the speed of the wind is 2.1428... mph.

I hope this helps!
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Answer by AnlytcPhil(1806)   (Show Source):
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Problem: Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and in the same amount of time runs 4 miles against the wind. What is the rate of the wind? Make this chart DISTANCE RATE TIME With wind Against wind >>...he runs 10 miles with the wind, and in the same amount of time runs 4 miles against the wind...<< Fill in the two distances DISTANCE RATE TIME With wind 10 Against wind 4 Let the rate of the wind be x miles per hour. So when running with the wind, his rate is increased by x mph. So we add x to his rate of 5mph and get 5+x mph, so fill that rate in: DISTANCE RATE TIME With wind 10 5+x Against wind 4 When running against the wind, his rate is decreased by x mph. So we subtract x from his rate of 5mph and get 5-x mph, so fill that rate in: DISTANCE RATE TIME With wind 10 5+x Against wind 4 5-x Now use TIME = DISTANCE/RATE to fill in the two times: DISTANCE RATE TIME With wind 10 5+x 10/(5+x) Against wind 4 5-x 4/(5-x) >>>...in the same amount of time...<< This says the two times are equal: 10 4 �� = �� 5+x 5-x Multiply thru by LCD = (5+x)(5-x) 10(5-x) = 4(5+x) 50 - 10x = 20 + 4x -14x = -30 x = (-30)/(-14) x = 15/7 or 2 1/7 mph Checking: When he runs with the wind he runs 5 + 2 1/7 mph or 7 1/7 mph or 50/7 mph for 10 miles. Since T = D/R, his time is 10/(50/7) = 7/5 hours When he runs against the wind he runs 5 - 2 1/7 mph or 2 6/7 mph or 20/7 mph for 4 miles. Since T = D/R, his time is 4/(20/7) = 7/5 hours So the times are equal, thus the answer is correct. Edwin AnlytcPhil@aol.com