Relative motion in three dimensions

[Kruz87]

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.

Homework 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)

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?

 

[AvroArrow]

The Attempt at a SolutionFor this problem, you need to use the equation Vpa=Vpb + Vba, where Vpa is the velocity of the plane as measured by the ground, Vpb is the velocity of the plane as measured by the wind, and Vba is the velocity of the wind as measured by the ground.Since we are given the velocity of the plane as measured by the ground (Vpa), 840 km/2 h = 420 km/h, and the velocity of the plane as measured by the wind (Vpb), 470 km/h, we can calculate the velocity of the wind as measured by the ground (Vba) using the equation:Vba = Vpa - Vpb = 420 km/h - 470 km/h = -50 km/h This means that the magnitude of the wind velocity is 50 km/h, and its direction is opposite of the direction of the plane's motion, which is 24.0° east of due north. Therefore, the direction of the wind velocity is 24.0° west of due north, or -24.0° relative to due west, where north of west is a positive angle and south of west is a negative angle.

 

[m2003]

I appreciate your attempt at solving this problem. However, I believe you may have made a mistake in your calculation of Vpw. The plane's velocity in relation to the wind should be equal to the airspeed of the plane (470 km/h) in the direction of the wind, which is 24.0° east of due north. This means that Vpw should be equal to 470 km/h*cos(24)i + 470 km/h*sin(24)j = 429.366i + 191.166j. Therefore, Vwg = Vpg - Vpw = (0-429.366)i + (420-191.166)j = -429.366i + 228.834j. The magnitude of Vwg is still sqrt(429.366^2 + 228.834^2) = 488.18 km/h and the angle is tan^-1(191.166/429.366) + 90 degrees = 28.07° west of due north. I believe this should be the correct answer, but please double check my calculations to be sure.