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Old Sep 7th 2010, 02:44 PM   #1
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Exclamation Simple reference frame problem

You are on an airplane traveling due east at 100 m/s with respect to the air. The air is moving with a speed 35 m/s with respect to the ground at an angle of 30 west of due north.

What is the speed of the plane with respect to the ground?
__________

So here's what Im doing:
Drew the whole right triangle thing to bring down air into its horizontal and vertical components. Vertical turned out to be 17.5m/s and horizontal 30.31m/s.

Then I broke up the speed of the plane into its horizontal and vertical components, but I used the same angle of 30 to get horizontal speed of 86.6 and vertical speed of 50. I'm pretty sure this part is wrong.

So to get the speed of the plane wrt ground, the equation is: V[p,g]=V[p,a] + V[a,g]
so v[p,g]=86.6+30.31=116.91. That was the wrong answer. I tried subtracting-wrong answer. I even used 60 degrees instead of 30 for breaking up the plane into vertical and horizontal, but that didnt work out either. I have no idea what Im doing.
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Old Sep 8th 2010, 11:16 AM   #2
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You're on the right track with respect to breaking the components of velocity into horizontal and vertical pieces. It helps to draw out a sketch - see below.

First, you gave the angle of the wind as 30 degrees west of north, which means it is 60 degrees from teh horizontal axis. The horizontal component of the wind is therefore -35m/s x cos(60) = -17.5 m/s, and the vertical component is 35 m/s x sin(60) = 30.31 m/s. So that was the first problem - you had these reversed. Note the minus sign for the horizontal piece - that's because it is the west (left). The horizontal component of the wind is +100 m/s, and the vertical component is 0 m/s - since the plane is heading due east. Now add the horoizontal pieces and then the vertical pieces:

V_h = 100 - 17.5 = 82.5m/s
V-v = 30.31 m/s

The total velocity is found using the Pythagorian theorem:

V_t = sqrt (V_h^2 + V_v^2) = sqty (82.5^2 + 30.31^2) = 87.9 m/s

Hope this helps.

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