# Intersection

• Jan 22nd 2007, 01:49 PM
symmetry
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.
• Jan 22nd 2007, 05:55 PM
Soroban
Hello, symmetry!

Quote:

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 $\displaystyle d$ between the cars as a function of time $\displaystyle t$.

Code:

            C    3 - 40t    B  40t  Q             * - - - - - - - - * - - - - *             |              *             |          *     2 - 30t |        * d             |    *             |  *           A*             |         30t |             |           P*

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

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

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

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

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

• Jan 22nd 2007, 05:59 PM
symmetry
ok
Like always, your replies are greatly appreciated as I study for my test.
• Jan 22nd 2007, 07:47 PM
AfterShock
Quote:

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,