Question #381be

1 Answer
Aug 26, 2017

Here's what's going on here.

Explanation:

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.