# Homework Help: Finding total distance on a position vs time graph

1. Mar 20, 2016

### physicsnerd26

I feel terrible for even having to ask this. I'm doing A2 at the moment, and we've gone through these sorts of lessons ever since my early AS days. I'm trying to review my notes cos exams are coming soon and I did some questions, and I'm apparently doing them wrong with what the answers are showing me at the back of the book.

Can anybody explain it to me? I'm talking about the kind of graph that goes from negative to positive.
Here's an example of a graph I'm talking about I found on the internet:

What I do is technically
A to B would be -> 10 x 2 = 20m
B to C would be -> (10 x 2)/2 = 10m
C to D would be 0
D to E would be - > (0.5 x -15)/2 = -3.75m, but it's distance so it's 3.75m
E to F would be 0
F to G would be -> (15 x 1)/2 =7.5m
G to H would be -> (2 x 15)/2 = 15m

so overall, I just add them all up so, 20+10+3.75+7.5+15 = 56.25m

Have I been doing it completely wrong?

2. Mar 20, 2016

### Staff: Mentor

I moved the thread to our homework section.
If you spend two seconds standing on a mark on the floor that says "10 meters", which distance did you cover during those two seconds?
If you spend one second running from a mark "-16 meter" to a mark "0" 16 meters away, what is the distance you covered?

It is a position versus time graph, not a speed versus time graph.

3. Mar 20, 2016

### physicsnerd26

So I should do position/time instead of multiplying them together?

Still 16, wouldn't it?

Yeah, I get confused between the two most of the time. Can you tell me if my answers above are correct?

4. Mar 20, 2016

### Staff: Mentor

The answer is much easier. Just check where you are when. If you have to move from position x to y, you have to move by ...
It does not matter how long that takes.
Right, but you calculated a different, wrong value.
They are not.

5. Mar 20, 2016

### Staff: Mentor

This makes no sense. Between A and B, the position didn't change. Also, the units don't make sense -- you're multiplying a position (in m.) by a time (in sec.) so you wouldn't get meters as a result. In fact, you don't get anything meaningful by multiplying the position by the time.