SOLUTION: a clclist rides 16 miles per hour on level ground on a still day. He finds that he finds that he rides 48 miles with the wind behind him in the same amount of time that he rides 16

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Question 31133This question is from textbook Intro Algebra
: a clclist rides 16 miles per hour on level ground on a still day. He finds that he finds that he rides 48 miles with the wind behind him in the same amount of time that he rides 16 miles into the wind. find the rate of the wind This question is from textbook Intro Algebra

Answer by mbarugel(146) About Me  (Show Source):
You can put this solution on YOUR website!
Hello!
We know that the cyclist can ride at 16 mph when there is no wind. Now, let's call X to the speed of the wind, measured in mph. When the wind is behind the cyclist, his final speed is his own speed (16 mph) plus the speed of the wind:
16+%2B+X
On the other hand, when he rides into the wind, his own speed is decreased by the speed of the wind. His final speed is:
16+-+X
We also know that he rides 48 miles with the wind in the same time that he rides 16 miles against the wind. This implies that his speed when going with the wind is 48/16 = 3 times greater than when he rides against the wind. Therefore, we get the equation:
16%2BX+=+3%2816-X%29
[his speed with the wind is 3 times his speed against the wind]
So now we just solve this simple equation for X:
16%2BX+=+48+-+3X
4X+=+48-16=32
X+=+8
The speed of the wind is 8 mph.

I hope this helps!
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