Question

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.

Answer 1

Answer:

$v_p=130\ miles/h$

$v_w=30\ miles/h$

Step-by-step explanation:

Lets call $v_p$ the the still-air speed of the plane and $v_w$ the speed of the wind. On the first part the speed of the plane relative to the ground will be $v_1=v_p-v_w$, and on the second part it will be $v_2=v_p+v_w$.

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:

$v_1=\frac{d}{t1}$

$v_2=\frac{d}{t2}$

Or:

$v_p-v_w=\frac{d}{t1}$

$v_p+v_w=\frac{d}{t2}$

Adding both equations:

$(v_p-v_w)+(v_p+v_w)=2v_p=\frac{d}{t1}+\frac{d}{t2}$

$v_p=\frac{1}{2}(\frac{d}{t1}+\frac{d}{t2})$

Which for our values is:

$v_p=\frac{1}{2}(\frac{400\ miles}{4h}+\frac{400\ miles}{2.5h})=130\ miles/h$

If instead of adding, we substact, we would have:

$(v_p+v_w)-(v_p-v_w)=2v_w=\frac{d}{t2}-\frac{d}{t1}$

$v_w=\frac{1}{2}(\frac{d}{t2}-\frac{d}{t1})$

Which for our values is:

$v_w=\frac{1}{2}(\frac{400\ miles}{2.5h}-\frac{400\ miles}{4h})=30\ miles/h$