Question

A plane travels speed at 225 mph in still air. Flying with tailwind, the plane is clocked over distance of 875 miles. Flying against headwind, it takes 1 hour longer to complete the return trip. What is the wind velocity?

Answer 1

Answer:

The wind speed is 28.466 mi/h

Step-by-step explanation:

Let's call Vs=225mi/h the plane speed in still air. Let's have X=875 mi the distance traveled

We'll also call Vw the wind speed. In the first flight, the plane goes with a speed of Vs+Vw.

The return trip is made flying against a headwind with a speed of Vs-Vw

The time taken to travel X miles with a tailwind is

$t_1=\frac{X}{Vs+Vw}$

The time taken to travel X miles with a headwind is

$t_2=\frac{X}{Vs-Vw}$

We know $t_1=t_2-1$ because the return trip is 1 hour longer. Then we have

$\frac{X}{Vs+Vw}=\frac{X}{Vs-Vw}-1$

Multiplying by (Vs+Vw)(Vs-Vw)

$X(Vs-Vw)=X(Vs+Vw)-(Vs^2-Vw^2)$

Replacing the values of X=875 and Vs=225 we reach a second-degree equation

$Vw^2+1750Vw-50625=0$

Which has the following roots:

Vw=28.466, Vw=-1778.466

We take the positive root and conclude

The wind speed is 28.466 mi/h

Note: We can easily check that the first time is 3.45h and the second time is 4.45h.