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Distance traveled by airplane 
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#1
Apr1010, 02:27 AM

P: 895

1. The problem statement, all variables and given/known data
An airplane leaves the airport when there is an Easterly wind (i.e a wind blowing from direction East) of speed 43 km/h. If the plane starts out directly above Taupo with an airspeed v=140 km/h, pointing in a direction θ = 26° East of North, how far is it from Taupo after 63 min of flying at constant altitude? (Ans: 134 km) 2. Relevant equations v=d/t 3. The attempt at a solution I found the x and y components of the velocity: V_{x}=140 cos 26 = 125.83 V_{y}=140 sin 26 = 61.37 Since the wind is blowing from the East at a speed of 43 km/h: V_{x}=125.83  43 = 82.83 km/h I could now use the formula v_{x}=d/t, but I don't think the problem is asking for just the displacement in the x direction. I'm very confused. Can anyone show me what I need to do? 


#2
Apr1010, 03:15 AM

P: 53

Lets look at exactly what the problem is asking:
"how far is it from Taupo after 63 min of flying at constant altitude?" Notice that it says how far. not how far east or west, just how far. So make a graph, draw a point at the origin, and a point where the plane is at after 63 minutes. The length of the line you draw between them is what you are trying to find. After seeing this it will be obvious that: [tex]L = \sqrt{d_{x}^{2} + d_{y}^{2}}[/tex] Where L is the length you are trying to find, [tex]d_{x}[/tex] is the distance in the x direction traveled by the plane, and [tex]d_{y}[/tex] is the distance traveled in the y direction. Try and take it from here. 


#3
Apr1010, 04:49 AM

P: 895

Okay, I drew the diagram and here are my calculations: V_{x}=140 cos 26 = 125.83 V_{y}=140 sin 26 = 61.37 d_{x}=v_{x}t=125.83 x 1.05 = 132.12 d_{y}=v_{y}t= 61.37 x 1.05 = 64.43 (63 Minutes = 1.05 Hours) [tex]L=\sqrt{(132.12)^2+(64.43)^2}=146.99[/tex] The correct answer to this problem is 134 km. Why do I not get this answer? And what about the Easterly wind, shouldn't we take that into account as well? 


#4
Apr1010, 06:18 AM

P: 11

Distance traveled by airplane
Did you draw the situation?
In my opinion in Vx should be sin, and Vy cos (pointing in a direction θ = 26° East of North). [tex]v = \sqrt{(v_w  v_x)^2 + v_y^2}[/tex] (the Pythagorean theorem) [tex]v_w[/tex]  speed of wind And then d=vt I got 135,45km and I think it's a fault of approximation. 


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