# Joe flies a plane 300 miles in 2 hours to a destination with the wind. On the return trip against the wind, he only flies 270 miles in 2 hours. Assuming wind speed is constant, what's the wind speed?

Airplane = 142.5 mph. Wind = 7.5 mph.

#### Explanation:

Let's think of the problem this way:

On the first leg of the trip , Joe flew 300 miles in 2 hours. We can then say that he flew:

$\left(300 \text{ miles")/(2 " hours}\right) = 150 m p h$

Since this was in the same direction of the wind, the plane was going the speed of the plane in still air plus the speed of the wind:

$\text{Airplane speed " + " wind speed} = 150 m p h$

On the return trip, Joe flew 270 miles in the same 2 hours. We can say that he flew:

$\left(270 \text{ miles")/(2 " hours}\right) = 135 m p h$

Since this was in the opposite direction of the wind, the plane was going the speed of the plane in still air minus the speed of the wind:

$\text{Airplane speed " - " wind speed} = 135 m p h$

Putting it together

We can add the two equations:

$\text{Airplane speed " + " wind speed} = 150 m p h$
ul("Airplane speed " - " wind speed" = 135 mph
$2 \times \text{Airplane speed} = 285 m p h$

Which means that $\text{Airplane speed} = \frac{285}{2} = 142.5 m p h$

The wind speed is then: $142.5 + \text{wind speed"=150=>"wind speed} = 7.5 m p h$

Apr 14, 2017

Wind speed: $w = 7.5 m p h$

Speed of plane in still air: $142.5 m p h$

#### Explanation:

Flying with the wind Joe flew $300 m$ in $2 h$
Flying against the wind Joe flew $270 m$ in $2 h$

Then we can equate his relative speeds (in still air) there and back to his present location, since the times are the same. We will use $w$ for the wind speed:

Outbound, Joe's air speed (no wind) is: $\frac{300 m p h}{2 h} - w$

The wind has to be subtracted because here it is adding to the speed of the plane in still air.

Inbound, Joe's air speed (no wind) is: $\frac{270 m p h}{2 h} + w$

The wind has to be added because here it is reducing the speed of the plane in still air.

Then: $\frac{300}{2} - w = \frac{270}{2} + w$

$\frac{300}{2} - \frac{270}{2} = w + w$

$300 - 270 = 2 \left(w + w\right)$

$30 = 4 w$

$w = 7.5 m p h$ which is the wind speed

Outbound, Joe's air speed (no wind) is: $\frac{300 m p h}{2 h} - w$

$\frac{300}{2} - 7.5 = 142.5 m p h$

Inbound, Joe's air speed (no wind) is: $\frac{270 m p h}{2 h} + w$

$\frac{270}{2} + 7.5 = 142.5 m p h$