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Relative Wind Problem (Modern Engineering Mathematics, 5th)

  • Thread starter SubZer0
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  • #1
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Homework Statement



A cyclist travelling east at 8 km/hr finds that the wind appears to blow directly from the north. On doubling his speed it appears to blow from the north-east.. Find the actual velocity of the wind.[/B]


Homework Equations



Wind(relative) = Wind(Actual) - Velocity(cyclist)

The Attempt at a Solution



I am completely stuck on this. I do know that there is a (16, 0) vector for the velocity of the cyclist, and a vector of -a(cos(45 deg), sin(45 deg)), which represents the relative, or perceived wind.

Slotting this into the relative wind equation,

Wind(r) = Wind(a) - Vel(cyclist)

-a*cos(45 deg) = Wx - 16
-a*sin(45 deg) = Wy

I'm not sure if this is even on the correct track.

Any trips highly appreciated.
 

Answers and Replies

  • #2
haruspex
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not sure if this is even on the correct track.
It is.
You just need to write the corresponding equations for the 8km/h case.
 
  • #3
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It is.
You just need to write the corresponding equations for the 8km/h case.
Thanks heaps for the reply, haruspex. I think I just needed that little confidence boost you gave me to help solve the problem.

So now I have the full set of equations:

-a*cos(45 deg) + 16 = Wx
-a*sin(45 deg) = Wy
Wx = 8
Wy = b

Slotting Wx into the first equation;

8 = 16 - a*cos(45 degrees)
a = 11.3137

Slotting a into (2), gives:

Wy = -11.3137*sin(45 degrees)
Wy = -7.999 (approx 8)

Therefore, actual wind velocity has vector of (8, -8). Magnitude, of sqrt(64+64) = 11.31km/hr

Thanks, haruspex!
 
  • #4
haruspex
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actual wind velocity has vector of (8, -8).
Looks right.
 

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