# Math Help - Distance Rate Time Word Problem

1. ## Distance Rate Time Word Problem

Im not quite sure how to set this word problem up:

Two boats leave simultaneously from the same port. One travels due north and the other travels due west. The northbound boat travels 5 mph faster than the eastbound boat, and after 2 hours the boats are 100 miles apart. Find the rate at which each boat is traveling.

Any help would be much appreciated

2. I would say it's a algebra problem, not a trigonometry problem.

Either way:

$\frac{100}{2} = x + (x+5)
$

$50 = 2x+5$

$2x = 45$

$x = 22.5$

the first speed would be 22.5 mph
the second would be that plus five so 27.5 mph

3. Hello, maryanna91!

Im not quite sure how to set this word problem up.
I'm not surprised, you've mixed two word problems.
Two boats leave simultaneously from the same port.
One travels due north and the other flies due west.
The northbound boat flies 5 mph faster than the eastbound boat,
and after 2 hours the planes are 100 miles apart.
Find the rate at which each boat is traveling.
I'll take a guess at what you meant . . .
Code:
                  o N
* ↑
100   *   ↑
*     ↑ 2(x+5)
*       ↑
*         ↑
W o ← ← ← ← ← o P
2x

Let $x$ = rate of the westbound boat (in mph).
Then $x+5$ = rate of the northbound boat.

They start together at P.

After 2 hours, the westbound boat travels $2x$ miles to $W.$
During the same 2 hours, the northbound boat travels $2(x+5)$ miles to $N.$

At that time, they are 100 miles apart: . $NW = 100$

Pythagorus says: . $(2x)^2 + [2(x+5)]^2 \:=\:100^2$

. . which simplifies to: . $2x^2 + 40x - 2475 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{-10 \pm\sqrt{19,\!900}}{4}$

. . The positive root is: . $x \;=\;32.76683995$

Therefore: . $\begin{Bmatrix}\text{westbound} &\approx& 32.8\text{ mph} \\ \text{northbound} &\approx& 37.8\text{ mph} \end{Bmatrix}$

4. Thanks a lot! Sorry for the mix up, it was much too early for math.