# Thread: Algebra - Wind problem

1. ## Algebra - Wind problem

I can't figure this problem out.

During a 3000 mile flight, a plane encountered a strong tail wind that increased its speed by 100 mi/hour. This increase of speed shortened the flying time by one hour. Find the speed of the plane relative to the ground.

2. Hello, SwissArmelle!

I'll do it the "standard" way", but we'll get a tricky equation . . .

During a 3000-mile flight, a plane encountered a strong tail wind
that increased its speed by 100 mi/hour.
This increase of speed shortened the flying time by one hour.
Find the speed of the plane relative to the ground.
We will use: .$\displaystyle \text{Distance} \:=\: \text{Speed} \times \text{Time}\quad\Rightarrow\quad t \:=\:\frac{d}{s}$

Let $\displaystyle S$ = speed of the plane (ground speed).

Normally, it would fly the 3000 miles at $\displaystyle S$ mph.
. . It would take: .$\displaystyle \frac{3000}{S}$ hours.

With the tailwind, its speed is $\displaystyle S + 100$ mph.
. . To fly 3000 miles, it took only: .$\displaystyle \frac{3000}{S + 100}$ hours.
And this is one hour less than its normal time.

There is our equation . . . . $\displaystyle \frac{3000}{S + 100} \:=\:\frac{3000}{S} - 1$

Multiply by $\displaystyle S(S+100)\!:\;\;3000S \:=\:3000(S + 100) -S(S + 100)$

This simplifies to the quadratic: .$\displaystyle S^2 + 100S - 30,000 \:=\:0$

. . which factors: .$\displaystyle (S - 500)(S + 600) \:=\:0$

. . and has roots: .$\displaystyle S \:=\:500,\:-600$

Therefore, the plane's ground speed is: $\displaystyle \boxed{500\:mph}$