Question #381be

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

`"1 min = 60 s"`

so

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

Now, notice that in `"600 s"`, the car travels a total distance of

`d = v * t`

`d = "15 m/"cancel("s") * 600cancel("s") = "9000 m"`

Assuming that this car is moving along the `x` axis, if you take `"0 m"` to be the origin of the axis, the initial position of the car will be at `-"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 `"200 m"` from its initial location to the town square and `"8800 m"` from the town square to its new position.

`"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!

`color(white)(aaaaaa)"start"color(white)(aaa)"town square"color(white)(aaaaaaaaaaa)"finish"`
`larrcolor(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)rarr`
`color(white)(aaaa)-200color(white)(aaaaaaa)0color(white)(aaaaaaaaaaaaaaa)+8800`

This is why the equation uses `-"200 m"` and not `+"200 m"` as the initial displacement of the car.