• Apr 8th 2008, 02:40 PM
lilikoipssn
Car A and Car B leave the origin at noon. Car A travels north at 66 mph and car B travels east at 112 mph.
(a) How far apart are they after 1 hour?
(b) How far apart are they after http://homework.math.ucsb.edu/webwor...5b59f2bc11.png hours?
(c) When are they 1300 miles apart?

I know (a) is 130 miles. But I can't seem to get b, and therefore i can't do c. If anyone can help I'd greatly appreciate it =)
• Apr 8th 2008, 02:59 PM
topher0805
What you are looking for is distance (D) as a function of time (t).

$\displaystyle D(t)$

Well, we know the equations that give us the distance of Car A and Car B from the origin:

$\displaystyle D_A(t)=66t$

$\displaystyle D_B(t)=112t$

And we know that to get the distance between the two we use Pythagoras Theorem.

So, we have that:

$\displaystyle D_{AB}(t)=\sqrt {(D_A)^2+(D_B)^2}$

We then add in our knowledge of $\displaystyle D_A$ and $\displaystyle D_B$ to get:

$\displaystyle D_{AB}(t)=\sqrt {(66t)^2+(112t)^2}$
• Apr 8th 2008, 03:25 PM
lilikoipssn
Thank you so much!
• Apr 8th 2008, 03:26 PM
Soroban
Hello, lilikoipssn!

Quote:

Car A and Car B leave the origin at noon.
Car A travels north at 66 mph and car B travels east at 112 mph.

(a) How far apart are they after 1 hour?
(b) How far apart are they after $\displaystyle t$ hours?
(c) When are they 1300 miles apart?

We can solve all the problems at once . . .
Code:

    P *       |  *       |    *  d   66t |        *       |          *       |              *       * - - - - - - - - *       O      112t

Car $\displaystyle A$ drives north at 66 mph.
. . In $\displaystyle t$ hours, it has gone $\displaystyle 66t$ miles to point $\displaystyle P$

Car $\displaystyle B$ drives east at 112 mph.
. . In $\displaystyle t$ hours, it has gone $\displaystyle 112t$ miles to point $\displaystyle Q.$

Their distance $\displaystyle d$ is the hypotenuse of right triangle $\displaystyle POQ.$

. . $\displaystyle d \;=\;\sqrt{(66t)^2 + (112t)^2} \;=\;\sqrt{16900t^2} \;=\;130t$ miles.

(a) When $\displaystyle t = 1\!:\;\;d \:=\:130(1) \:=\:130$ miles.

(b) At time $\displaystyle t$, their distance is: .$\displaystyle d \:=\:130t$ miles.

(c) If $\displaystyle d = 1300$, we have: .$\displaystyle 130t \:=\:1300 \quad\Rightarrow\quad t \:=\: 10$ hours.