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?

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Answer 1 · 2015-10-08T12:43:09

The speed of the plane is 225 mph.

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

${(s/(v+25)=4),(s/(v-25)=5):}$

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:

${(s=4*(v+25)),(s=5*(v-25)):}$

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

$4*(v+25)=5*(v-25)$

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

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

Check:

First we calculate the distance:

$s=4*(225+25)=4*250=1000$

Next we check the times:

a) with the wind

$t=s/(v+25)=1000/(225+25)=1000/250=4$

b) against the wind

$t=s/(v-25)=1000/(225-25)=1000/200=5$

Answer 2 · 2015-10-08T12:43:15

I found $225mph$

Call the speed of the plane (in still air) $s_p$ ; you get from speed=distance/time:
$s_p+25=d/4$
and:
$s_p-25=d/5$

from the first equation:
$d=4(s_p+25)$
substitute into the second:
$s_p-25=(4(s_p+25))/5$
$5s_p-125=4s_p+100$
$s_p=225mph$