the distance

• Sep 18th 2010, 04:47 PM
rai2003
the distance
A boat with an ill passenger is 7 1/2 mi north of a straight coastline which runs east and west. A hospital on the coast is 60 miles from the point on shore south of the boat. If the boat starts toward shore at 15 mph at the same time an ambulance leaves the hospital at 60 mph and meets the ambulance, what is the total distance (to the nearest 0.5 mile) traveled by the boat and the ambulance?
• Sep 18th 2010, 06:43 PM
Soroban
Hello, rai2003!

There's something missing from your problem.

Quote:

A boat with an ill passenger is 7 1/2 mi north of a straight coastline which runs
east and west. A hospital on the coast is 60 miles from the point on shore south
of the boat. If the boat starts toward shore at 15 mph at the same time
an ambulance leaves the hospital at 60 mph and meets the ambulance
in the least amount of time, what is the total distance (to the nearest 0.5 mile)
traveled by the boat and the ambulance?

Code:

    B o       |  *      ___________       |    *  √x^2 + 7.5^2   7.5 |        *       |          *       |              *     S o-----------------o---------------o       : - - -  x  - - - P - - 60-x  - - H       : - - - - - - - 60  - - - - - - - :

The boat is at $\,B$, 7.5 miles from shore: . $BS = 7.5$

The hospital is at $\,H$: . $SH = 60$

The boat heads for point $\,P$ at 15 mph: . $SP = x$

Its distance $BP$ is the hypotenuse of a right triangle.
. . Hence: . $BP \,=\,\sqrt{x^2 + 7.5^2}$ miles.

At 15 mph, this will take: . $\dfrac{\sqrt{x^2+7.5^2}}{15}$ hours.

At the same time, the ambulance goes from $\,H$ to $\,P$ at 60 mph.

Its distance is: . $HP \,=\,60-x$ miles.

At 60 mph, this will take: . $\dfrac{60-x}{60} \:=\:1 - \dfrac{x}{60}$ hours.

The total time is: . $T \;=\;\dfrac{(x^2+7.5^2)^{\frac{1}{2}}}{15} + 1 - \dfrac{x}{60}$

And that is the function we must mininize.

Once we find $\,x$, we can determine the distances travelled.

• Sep 20th 2010, 03:05 PM
rai2003
There is not anything missing from my question. If you like you can go to this link to check the question http://www.amatyc.org/SML/SMLTestFall2003.pdf. I try your solution to find the answer but look like it does not work please can you try to solve it again. The answer for this question is 62.5. Thanks
• Sep 20th 2010, 06:01 PM
Soroban
Hello, rai2003!

Okay, I get it now.

Silly me, I assumed it was a "minimnum time" problem . . . I was wrong.

Use my original diagram . . .

The boat goes $\sqrt{x^2 + 56.25}$ miles at 15 miles per hour.

. . This takes: . $\dfrac{\sqrt{x^2+56.25}}{15}$ hours.

The ambulance goes $60-x$ miles at 50 miles per hour.

. . This takes: . $\dfrac{60-x}{60}$ hours.

The boat and the ambulance meet at point $\,P$ at the same time.

. . Hence, the two times are equal.

We have: . $\dfrac{\sqrt{x^2+56.25}}{15} \;=\;\dfrac{60-x}{60} \quad\Rightarrow\quad 4\sqrt{x^2+56.25} \;=\;60 - x$

Square both sides: . $16(x^2+56.25) \;=\;(60-x)^2$

. . $16x^2 + 900 \;=\;3600 - 120x + x^2 \quad\Rightarrow\quad 15x^2 + 120x - 2700 \:=\:0$

. . $x^2 + 8x - 180 \:=\:0 \quad\Rightarrow\quad (x-10)(x+18) \:=\:0$

. . $x \:=\:10,-18$

Hence: . $x \;=\;SP \;=\;10$ miles.

The boat travelled: . $BP \;=\;\sqrt{10^2 + 56.25^2} \:=\:\sqrt{156.25} \;=\;12.5$ miles.

The ambulance travelled: . $HP \;=\;60 - 10 \:=\: 50$ miles.

Their total distance was: . $12.5 + 50 \;=\;62.5$ miles.
• Sep 21st 2010, 01:40 PM
rai2003
thanks