Calculating Flight Paths

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1. Question

QuestionQuestion A pilot is flying from Daytona Beach to Prescott. The two cities are 1874 miles away from one another, and the angle from Daytona Beach to Prescott is W28N. At takeoff, there is a 12 mph wind blowing N37E that blows constantly throughout the entire 4-hour flight. At what angle should the pilot leave Daytona Beach to fly in a straight line and land in Prescott?

Flight Path Calculation Diagram

Figure 1: Flight Path Calculation Diagram

2. Solution

We call the velocity of the plane relative to the ground VgV_g. Cause two cities are 1874 miles away from one another and it took an entire 4-hour flight:

Vg=18744=468.5(miles/hour)\Rightarrow V_g = \dfrac{1874}{4} = 468.5 (miles/hour)

The angle from Daytona Beach to Prescott is W28N=152(degrees)W28N = 152 (degrees) in the X-Y coordinate, which means it is in the 2nd quadrant:

Vgx=468.5cos(28)413.66Vgy=468.5sin(28)219.95\begin{align*} V_gx&=-468.5\cos(28)\approx -413.66 \\ V_gy&=468.5\sin(28) \approx 219.95 \\ \end{align*}

We call the velocity of the wind Vw=12(miles/hour)V_w = 12(miles/hour). Cause wind blowing N37E=233(degrees)N37E = 233 (degrees) in the XYX-Y coordinate, which means it is in the 3rd quadrant:

Vwx=12sin(37)7.22Vwy=12cos(37)9.58\begin{align*} V_wx&= -12\sin(37) \approx -7.22 \\ V_wy &= -12\cos(37) \approx -9.58 \end{align*}

We call vector Vp\vec{V_p} the way the pilot leaves Daytona Beach to fly in a straight line and land in Prescott.

Vgx=Vpx+VwxVpx=VgxVwx=413.66(7.22)=406.44Vgy=Vpy+VwyVpy=VgyVwy=219.95(9.58)=229.53\begin{align*} V_gx&=V_px + V_wx & \Rightarrow V_px &= V_gx - V_wx = -413.66-(-7.22)=-406.44 \\ V_gy&=V_py + V_wy & \Rightarrow V_py &= V_gy - V_wy = 219.95-(-9.58)=229.53 \\ \end{align*}

Cause Vpx<0V_px < 0 and Vpy>0V_py > 0 so that means Vp\vec{V_p} is in the 2nd quadrant. We call the angle between Vp\vec{V_p} and the X-axis α\alpha (where is the angle from the West to the North):

tan(α)=VpyVpx=229.53406.440.565\tan(\alpha) = \left| \dfrac{V_py}{V_px}\right|=\left| \dfrac{229.53}{-406.44}\right| \approx 0.565 α=arctan(0.565)29.46(degrees)\Rightarrow \alpha = \arctan(0.565) \approx 29.46 (degrees)

Thus, the pilot should leave Daytona Beach at an angle of W29.46NW29.46N to fly in a straight line and land in Prescott. That means the plane flew northeast at an angle of 1.46(degrees)1.46 (degrees).

3. Conclusion

The pilot should leave Daytona Beach at an angle of W29.46NW29.46N to fly in a straight line and land in Prescott. That means the plane flew northeast at an angle of 1.46(degrees)1.46 (degrees).