A light plane attains an airspeed of 470 km/h. The pilot sets out for a destination 840 km due north but discovers that the plane must be headed 24.0° east of due north to fly there directly. The plane arrives in 2.00 h. What were the (a) magnitude and (b) direction of the wind velocity? Give the direction as an angle relative to due west, where north of west is a positive angle, and south of west is a negative angle.(adsbygoogle = window.adsbygoogle || []).push({});

2. Relevant equations

(relative motion in two dimensions)

Vpa= Vpb + Vba (the velocity of Vpa of P as measured by A is equal to the velocity Vpb of P as measured by B plus the velocity of Vba measured by A)

3. The attempt at a solution

Okay, so here was my valiant but wrong attempt. So first, I drew and labeled my diagram using the three following vectors: (1) Vpg: The plane's velocity in relation to the ground in which I had it angeled 24.0° east of due north (forms hypotenuse of triangle). (2) Vpw: The plane's velocity in relation to the wind in which I had it directly on the y-axis (forms adjacent of triangle ). And (3) Vwg: The wind's velocity in relation to the ground which I had angled westbound to cause the plane to go straight (forms opposite side of 24° angle in triangle). Next, I then made a relationship among the vectors using the triangle that I was able to draw: Vpg =Vpw + Vwg or Vwg= Vpg-Vpw. Where Vpg= 840km/2h= 0i + 420j (b/c only in y direction) and Vpw=470km/h*cos(24)i + 470km/h*sin(24)j= 429.366i + 191.166j. Therefore, Vwg=(0-429.366)i+(420-191.166)j= -429.66i + 228.833j. The magnitude is sqrt(429.66^2 + 228.833j^2) And the angle is tan^-1(191.166/429.366) +90 degrees. It seemed so right, where did I go wrong???

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# Homework Help: Relative motion in three dimensions

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