# Thread: Cars

1. ## 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.

2. Hello, blueridge!

Did you make a sketch?

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

#### Search Tags

cars 