1. ## Intersection

Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 mph. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 mph. Express the distance d between the cars as a function of time t.

2. Hello, symmetry!

Two cars are approaching an intersection.
One is 2 miles south of the intersection and is moving at 30 mph.
The other car is 3 miles east of the intersection and is moving at 40 mph.
Express the distance $d$ between the cars as a function of time $t$.
Code:
            C     3 - 40t     B   40t   Q
* - - - - - - - - * - - - - *
|              *
|           *
2 - 30t |        * d
|     *
|  *
A*
|
30t |
|
P*

The first car starts at $P$ and drive north to the intersection $C$.
In $t$ hours, it has gone $30t$ miles to point $A$.
. . Hence, $AC \:=\: 2 - 30t$

The other car starts at $Q$ and drives west to $C$.
In $t$ hours, it has gone $40t$ miles to point $B$.
. . Hence, $BC \:=\:3 - 40t$

Using Pythagorus, the distance between them is: . $d^2\;=\;(2-30t)^2 + (3-40t)^2$

Simplifying, we get: . $d^2\;=\;2500t^2 - 360t + 13$

Therefore: . $\boxed{d \;=\;\sqrt{2500t^2 - 360t + 13}}$

3. ## ok

Like always, your replies are greatly appreciated as I study for my test.

4. Originally Posted by symmetry
Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 mph. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 mph. Express the distance d between the cars as a function of time t.
We have D = r*t

Distance of the slow car moving at 30 mph is 30t
Distance of the fast car moving at 40 mph is 40t

Now, this problem is practically asking you to use the pythag. theorem.

a^2 + b^2 = c^2

= sqrt(40t^2 + 30t^2)

= sqrt(1600t^2 + 900t^2)

= sqrt(2500t^2)

And thus,