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View Full Version : Relative Velocity as far as I can tell


merlinMan
Sep12-04, 07:25 PM
Before I get to the problem, Id just like to say a big thank you to all of you who help out and answer all of our questions. While many of them appear basic, to us they are daunting. It's nice to have a place to turn to when your professor barely speaks english and isn't a whole lot of help explaining the problem. So, many thanks to all of you who set aside your time to help the less knowledgable.

Down to the problem. City A lies due West of city B. A pilot normally makes the 5310km ROUND trip in 6.6hours. One day there is a steady wind blowing from west to east and it takes the pilot 6.7 hours to make the round trip. How fast is the wind blowing.

I've spent a while on this problem and can't seem to get it.

It makes sense to me that if your wind slows you one way, it would speed you up the other way, thereby canceling out the winds effects.

Not having any formulas to really draw upon, I came up with one for the planes velocity. 804.55 normally, so it would be 804.55 - w in one direction and 804.55 + w in the opposite. Doing vector addition canceled out the W. A little lost.

Doc Al
Sep12-04, 07:52 PM
It makes sense to me that if your wind slows you one way, it would speed you up the other way, thereby canceling out the winds effects.
Don't guess... figure it out!

Not having any formulas to really draw upon, I came up with one for the planes velocity. 804.55 normally, so it would be 804.55 - w in one direction and 804.55 + w in the opposite.
Excellent. Now figure out the time for each part of the trip. Use T = D/V. You know what the total time for the round trip must be.
Doing vector addition canceled out the W.
You can't add these "vectors" since they refer to different things! One applies in one direction, the other applies in the other direction. You can't just "average" them, since they apply for different times. (As you'll see when you figure it out!)

merlinMan
Sep12-04, 08:03 PM
See, I tried Time = Distance/Velocity. Then I added the two times together, Distance/(Velocity West) + Distance/(Velocity East) = 6.7

From there I got a equation I really couldn't work out. Is that how its done? I got so muddled down in the equation I figured I was doing it wrong. If not, i'll go back a muddle a little more.

Leong
Sep12-04, 08:59 PM
\frac{2655}{804.55+w}+\frac{2655}{804.55-w}=6.7

\frac{(804.55-w)+(804.55+w)}{804.55^2-w^2}=\frac{6.7}{2655}

I think you can continue from here...

merlinMan
Sep12-04, 09:56 PM
Ahh doh! stupid, thanks for your help