# Thread: distance rate time problem

1. ## distance rate time problem

I can't figure out
plane travels 680 mi one way with the wind at 300 mph plus tailwind
plane traves 520 mi back against the wind at 300 mph less headwind
the time to go and return are equal

I keep getting 300 as wind but that can't be

Thank you

2. ## since Distance = Rate x Time

Originally Posted by esther
I can't figure out
plane travels 680 mi one way with the wind at 300 mph plus tailwind
plane traves 520 mi back against the wind at 300 mph less headwind
the time to go and return are equal

I keep getting 300 as wind but that can't be

Thank you
since time is = then we have
let w = wind

since Distance = Rate x Time

$\displaystyle \frac{D_1}{R_1}= \frac{D_2}{R_2}$

$\displaystyle \frac{680m}{300mph - w} = \frac{520m}{300mph + w}$

$\displaystyle 680(300+w) = 520(300 - w)$

$\displaystyle 204000 + 680w = 156000 -520w$

$\displaystyle 48000 = -1200w$

$\displaystyle |- 40| = w = 40mph$

3. Hello, esther!

A plane travels 680 mi one way with the wind at 300 mph plus tailwind.
The plane travels 520 mi back against the wind at 300 mph less headwind.
The time to go and return are equal
Find the rate of the wind.

Let $\displaystyle x$ = rate of the wind.

With the wind, the plane flies 680 miles at $\displaystyle 300+x$ mph.
. . This takes: .$\displaystyle \frac{680}{300+x}$ hours.

Against the wind, the plane flies 520 miles at $\displaystyle 300-x$ mph.
. . This takes: .$\displaystyle \frac{520}{300-x}$ hours.

Since the times are equal, we have: .$\displaystyle \frac{680}{300+x} \;=\;\frac {520}{300-x}$

Divide by 40: .$\displaystyle \frac{17}{300+x} \:=\:\frac{13}{300-x}$

Cross-multiply: .$\displaystyle 5100 - 17x \:=\:3900 + 13x \quad\Rightarrow\quad 30x \:=\:1200\quad\Rightarrow\quad x \:=\:40$

The rate of the wind is 40 mph.