# Cars

• Jun 13th 2007, 01:40 PM
blueridge
Cars
Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 miles per hour. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 miles per hour. Express the distance d between the cars as a function of time t.

HINT GIVEN:

At t = 0, the cars are 2 miles south and 3 miles east of the intersection, respectively.
• Jun 13th 2007, 06:59 PM
Soroban
Hello, blueridge!

Did you make a sketch?

Quote:

Two cars are approaching an intersection.
One is 2 miles south of the intersection $\displaystyle X$ and is moving north at 30 mph.
The other car is 3 miles east of $\displaystyle X$ and is moving west at 40 mph.
Express the distance $\displaystyle d$ between the cars as a function of time $\displaystyle t$.

Code:

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

The first car starts at $\displaystyle P$, 2 miles south of $\displaystyle X\!:\;\;PX=2$
In $\displaystyle t$ hours, it has moved $\displaystyle 30t$ miles north to point $\displaystyle A.$
. . Hence: .$\displaystyle AX \:=\:2-30t$

The second car starts at $\displaystyle Q$, 3 mile east of $\displaystyle X\!:\;QX = 3$
In $\displaystyle t$ hours, it has moved $\displaystyle 40t$ miles west to point $\displaystyle B.$
. . Hence: .$\displaystyle BX \:=\:3-40t$

From right triangle $\displaystyle AXB$, we have: .$\displaystyle d^2 \:=\:AX^2 + BX^2$

So we have: .$\displaystyle d \;=\;\sqrt{(2-30t)^2 + (3-40t)^2}$

. . which simplifies to: .$\displaystyle d \;=\;\sqrt{2500t^2 - 360t +13}$