# An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. How do you find the speed of the airplane in still air?

Oct 8, 2015

The speed of the plane is 225 mph.

#### Explanation:

To solve this task you can use the following system of equations:

$\left\{\begin{matrix}\frac{s}{v + 25} = 4 \\ \frac{s}{v - 25} = 5\end{matrix}\right.$

Both equations come straight from the laws of cinematics (time equals distance divided by speed)

If you multiply both equations by the denominators you will get:

$\left\{\begin{matrix}s = 4 \cdot \left(v + 25\right) \\ s = 5 \cdot \left(v - 25\right)\end{matrix}\right.$

So you can write only one equation with the unknown $v$:

$4 \cdot \left(v + 25\right) = 5 \cdot \left(v - 25\right)$

$4 v + 100 = 5 v - 125$
$5 v - 4 v = 125 + 100$
$v = 225$

Answer: The speed of the plane without the wind is $225$ mph.

Check:

First we calculate the distance:

$s = 4 \cdot \left(225 + 25\right) = 4 \cdot 250 = 1000$

Next we check the times:

a) with the wind

$t = \frac{s}{v + 25} = \frac{1000}{225 + 25} = \frac{1000}{250} = 4$

b) against the wind

$t = \frac{s}{v - 25} = \frac{1000}{225 - 25} = \frac{1000}{200} = 5$

Oct 8, 2015

I found $225 m p h$

#### Explanation:

Call the speed of the plane (in still air) ${s}_{p}$; you get from speed=distance/time:
${s}_{p} + 25 = \frac{d}{4}$
and:
${s}_{p} - 25 = \frac{d}{5}$

from the first equation:
$d = 4 \left({s}_{p} + 25\right)$
substitute into the second:
${s}_{p} - 25 = \frac{4 \left({s}_{p} + 25\right)}{5}$
$5 {s}_{p} - 125 = 4 {s}_{p} + 100$
${s}_{p} = 225 m p h$