Question

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?

Answer 1

The speed of the plane is 225 mph.

Explanation:

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

I found `225mph`

Explanation:

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`