Position-time graph made with info from a velocity time graph

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SUMMARY

The discussion focuses on converting velocity-time graphs into position-time graphs using the formula distance = velocity × time. The user correctly calculates the position at various time intervals based on given velocities of 70 km/h, 90 km/h, and 80 km/h. They also inquire about calculating distance for different segments of a trip and understanding the distinction between distance and displacement. The user concludes with a comparison of the calculated distance and the actual displacement between Bradford and Milton, which is 71.6 km.

PREREQUISITES
  • Understanding of basic kinematics principles
  • Familiarity with velocity and speed concepts
  • Ability to perform unit conversions (e.g., minutes to hours)
  • Knowledge of graph interpretation, specifically position-time and velocity-time graphs
NEXT STEPS
  • Learn how to derive position-time graphs from velocity-time graphs
  • Study the differences between distance and displacement in physics
  • Explore the concept of average velocity and its calculation
  • Investigate how to apply kinematic equations in various motion scenarios
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Students studying physics, particularly those learning about motion, kinematics, and graph interpretation. This discussion is also beneficial for educators seeking to clarify these concepts for their students.

jesstryin
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Homework Statement



I have been given three velocity-time graphs, and have been asked to use the data from these to create position-time graphs. Each graph represent a route that generally goes south. Changes in direction to go east or west and gradual changes in velocity and time to stop at intersections, and so on, have not been included.

To explain the first velocity/time graph:

70km/h (S) traveled for 5 min, then in no time it jumps to 90km/h (S) from t=5 to t=35, then in no time it switches to 80km/h (S) from t=35 min to t=65 min

Homework Equations



I know that distance= time X speed, and velocity is change in position/change in time, so does that mean I use change in position = velocity X change in time?

The Attempt at a Solution



5min = 0.083 hours

V X T = 70km/h (s) X 0.083 h = 5.81 km
So, does this mean that at five minutes on the position-time graph, I mark down 5.81 km as the position? And for the second one:

30 min (from t=5 to t=35) = 0.5 hour
90km/h (s) X 0.5h = 45 km

So, do I mark down the first position as 5.81km and the second position as 45km?

If I did this wrong, please tell me the correct equation and show me how to complete this problem so I can learn, I have been trying to teach myself physics and it is hard, so being shown how to complete this problem would help
 
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Yes, that looks right. Another thing to think about when doing these graphs is that since velocity is distance over time, then if you have a position vs. time graph, the slope will be the velocity (i.e. rise over run = position over time). So for example, using the first velocity you could say that the position time graph will have a slope of 70 km/h from 0 to 5 minutes (watch out for units). Doing your calculations gives you the same answer -- this is just another way to think about these problems.
 
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Thanks! I am done making the position-time graphs but need help with the rest of the question. They ask me to record the distance for each route. I know distance = speed X time, but I only have the time, not the speed (the velocity is different from speed I think). Can you tell me what equation to use to record the distance, each route has different sections of the trip that are at different constant velocities.

Also, the last part of the question asks me to measure the displacement from Bradford to Milton using a map, then compare the difference between the distances traveled for each route with the displacement I find. I looked it up, and found that the exact distance is 71.6 km, (driving distance is 85 km according to the same map). Would this be the displacement, or am I doing something wrong?

Thanks again for your help I appreciate it :)
 

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