Finding total distance on a position vs time graph

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SUMMARY

The discussion centers on calculating total distance from a position versus time graph, specifically addressing common misconceptions among students. The user initially miscalculates distances by multiplying position values with time, leading to incorrect results. Key corrections emphasize that distance should be determined by the change in position, not by multiplying position by time. The correct approach involves simply assessing the difference in positions at given time intervals.

PREREQUISITES
  • Understanding of position versus time graphs
  • Basic knowledge of distance calculation
  • Familiarity with units of measurement (meters, seconds)
  • Concept of displacement versus distance
NEXT STEPS
  • Review the principles of position versus time graphs
  • Study the differences between distance and displacement
  • Practice calculating distance using various position-time graph examples
  • Learn about the implications of speed versus time graphs
USEFUL FOR

Students preparing for physics exams, educators teaching kinematics, and anyone seeking to clarify concepts related to motion and graph interpretation.

physicsnerd26
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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:
f71554f3-d21e-4a7d-986f-ea89e5208993.gif


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?
 
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I moved the thread to our homework section.
physicsnerd26 said:
A to B would be -> 10 x 2 = 20m
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.
 
mfb said:
I moved the thread to our homework section.

Sorry about that. And thanks.

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?

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

If you spend one second running from a mark "-16 meter" to a mark "0" 16 meters away, what is the distance you covered?
Still 16, wouldn't it?

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

Yeah, I get confused between the two most of the time. Can you tell me if my answers above are correct?
 
physicsnerd26 said:
So I should do position/time instead of multiplying them together?
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.
physicsnerd26 said:
Still 16, wouldn't it?
Right, but you calculated a different, wrong value.
physicsnerd26 said:
Yeah, I get confused between the two most of the time. Can you tell me if my answers above are correct?
They are not.
 
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physicsnerd26 said:
A to B would be -> 10 x 2 = 20m
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.
 
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