Question 381be

Aug 26, 2017

Here's what's going on here.

Explanation:

That's just the duration of the motion expressed in seconds.

$\text{1 min = 60 s}$

so

10 cancel("min") * "60 s"/(1cancel("min")) = "600 s"#

Now, notice that in $\text{600 s}$, the car travels a total distance of

$d = v \cdot t$

$d = \text{15 m/"cancel("s") * 600cancel("s") = "9000 m}$

Assuming that this car is moving along the $x$ axis, if you take $\text{0 m}$ to be the origin of the axis, the initial position of the car will be at $- \text{200 m}$ because the car is starting West of the town square and moving East, i.e. to the left of the origin and moving to the right.

The car will travel $\text{200 m}$ from its initial location to the town square and $\text{8800 m}$ from the town square to its new position.

$\text{initial position " stackrel(color(white)(acolor(blue)("200 m")aaa))(->)" town square " stackrel(color(white)(acolor(blue)("8800 m")aaa))(->) " final position}$

Using the $x$ axis for help, this is equivalent to--the diagram is not to scale!

$\textcolor{w h i t e}{a a a a a a} \text{start"color(white)(aaa)"town square"color(white)(aaaaaaaaaaa)"finish}$
$\leftarrow \frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}} \rightarrow$
$\textcolor{w h i t e}{a a a a} - 200 \textcolor{w h i t e}{a a a a a a a} 0 \textcolor{w h i t e}{a a a a a a a a a a a a a a a} + 8800$

This is why the equation uses $- \text{200 m}$ and not $+ \text{200 m}$ as the initial displacement of the car.