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 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 =)

2. What you are looking for is distance (D) as a function of time (t).

$D(t)$

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

$
D_A(t)=66t$

$
D_B(t)=112t$

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

So, we have that:

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

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

$D_{AB}(t)=\sqrt {(66t)^2+(112t)^2}$

3. Thank you so much!

4. Hello, 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 $t$ hours?
(c) When are they 1300 miles apart?
We can solve all the problems at once . . .
Code:
    P *
|  *
|     *   d
66t |        *
|           *
|              *
* - - - - - - - - *
O      112t

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

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

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

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

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

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

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