An airplane makes a 400-mile trip against a head wind in 4 hours. The return trip takes 2.5 hours, the wind now being a tall wind. If the plane maintains a constant speed with respect to still air, and the speed of the wind is also constant and does not vary, find the still-air speed of the plane and the speed of the wind.

Accepted Solution

Answer:[tex]v_p=130\ miles/h[/tex][tex]v_w=30\ miles/h[/tex]Step-by-step explanation:Lets call [tex]v_p[/tex] the the still-air speed of the plane and [tex]v_w[/tex] the speed of the wind. On the first part the speed of the plane relative to the ground will be [tex]v_1=v_p-v_w[/tex], and on the second part it will be [tex]v_2=v_p+v_w[/tex].We know that the distance d=400 miles on the first and second parts are the same, so by the definition of velocity we will have:[tex]v_1=\frac{d}{t1}[/tex][tex]v_2=\frac{d}{t2}[/tex]Or:[tex]v_p-v_w=\frac{d}{t1}[/tex][tex]v_p+v_w=\frac{d}{t2}[/tex]Adding both equations:[tex](v_p-v_w)+(v_p+v_w)=2v_p=\frac{d}{t1}+\frac{d}{t2}[/tex][tex]v_p=\frac{1}{2}(\frac{d}{t1}+\frac{d}{t2})[/tex]Which for our values is:[tex]v_p=\frac{1}{2}(\frac{400\ miles}{4h}+\frac{400\ miles}{2.5h})=130\ miles/h[/tex]If instead of adding, we substact, we would have:[tex](v_p+v_w)-(v_p-v_w)=2v_w=\frac{d}{t2}-\frac{d}{t1}[/tex][tex]v_w=\frac{1}{2}(\frac{d}{t2}-\frac{d}{t1})[/tex]Which for our values is:[tex]v_w=\frac{1}{2}(\frac{400\ miles}{2.5h}-\frac{400\ miles}{4h})=30\ miles/h[/tex]