# An airplane took 3 hours to fly 600 miles against a headwind. The return trip with the wind took 2 hours. Find the speed of the plane in still air and the speed of the wind?

Nov 14, 2017

Speed of the plane: 250 mph
Speed of the wind: 50 mph

#### Explanation:

Let p = the speed of the plane
and w = the speed of the wind

It takes the plane 3 hours to go 600 miles when against the headwind and 2 hours to go 600 miles with the headwind. So we set up a system of equations.

$\frac{600 m i}{3 h r} = p - w$

$\frac{600 m i}{2 h r} = p + w$

Solving for the left sides we get:

200mph = p - w

300mph = p + w

Now solve for one variable in either equation. I'll solve for x in the first equation:

200mph = p - w

p = 200mph + w

Now we can substitute the x that we found in the first equation into the second equation so we can solve for w:

300mph = (200mph + w) + w

Combine like terms:
300mph = 200mph + 2w

Subtract 200mph on both sides:
100mph = 2w

Divide by 2:
50mph = w

So the speed of the wind is 50mph.
Now plug the value we just found back in to either equation to find the speed of the plane, I'll plug it into the first equation:

200mph = p - 50mph

250mph = p

So the speed of the plane in still air is 250mph.

Nov 14, 2017

Ahhh, yes. Relative wind problems. There was a time when I had to do these in my head, while flying the plane.

Anyway, you have 2 equations in 2 unknowns. Let p be the plane's velocity, and w be the wind speed.

And remember, d=rt, where d = distance, r = rate, and t = time.

This lets you write your 2 equations. The rate on the upwind leg is
$p - w$, and the rate going downwind is $p + w$

so you can write:
$600 = \left(p - w\right) 3$ , and:
$600 = \left(p + w\right) 2$

or:

$3 p - 3 w = 600$
$2 p + 2 w = 600$

Let's take the second equation, and write w as a function of p.

$2 w = 600 - 2 p$
$w = 300 - p$

...now substitute this back in the first equation:

$3 p - 3 \left(300 - p\right) = 600$
$3 p - 900 + 3 p = 600$
$6 p = 1500$
$p = 250$

now, go back to the derivation we had for w:

$w = 300 - 250$
$w = 50$

Once again, using $d = r t$, $600 = 300 \left(t\right)$, so t = 2, which matches what we have in the problem statement: flying downwind takes 2 hours.
$600 = 200 \left(t\right)$, so $t = \frac{600}{200} = 3$, so upwind takes 3 hours, which checks out, too.