Given position vs time graph, find velocity?

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Homework Help Overview

The discussion revolves around analyzing a position vs. time graph to determine the velocity of a bicyclist at specific time intervals. Participants are exploring the relationship between position changes and time to calculate velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the slope of the graph to find velocity, questioning the correct interpretation of rise over run. There are attempts to clarify the calculation of distance traveled over time and how to apply this to different time intervals.

Discussion Status

Some participants provide guidance on calculating velocity by emphasizing the importance of understanding the graph's position values at different times. Multiple interpretations of the graph's behavior at various time points are being explored, particularly regarding stationary and negative velocities.

Contextual Notes

Participants are navigating the implications of the graph's slope and the significance of initial position values. There is a focus on how to handle time intervals where the position does not change, as well as the transitions between positive and negative velocities.

skysunsand
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Homework Statement



Given the graph in the uploaded screenshot, what is the bicyclist's velocity at t=10s?

Homework Equations



I have been trying to do a slope equation? Rise/run.

The Attempt at a Solution



Tried to do slope equation. Looks like 75/10, which would be 7.5, but that answer is incorrect. Why? It wants to two sig figs, and that is two sig figs...I also have no clue what to do if t= anything else, like 25 or something. Is there a different equation for that? I'm so lost. :(
 

Attachments

  • Screen shot 2011-08-23 at 10.10.50 PM.png
    Screen shot 2011-08-23 at 10.10.50 PM.png
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rise over run wouldn't be 75/10. it starts at 50 and then moves to 75. so the change in distance is 75-50=25. so rise over run is 25/10
 
remember speed is distance traveled/time taken. when time= 0, its position isn't zero like you assumed. It's position is 50. And then in 10 seconds its position is 75. So the distance traveled is 25. So s=d/t, where d=25 and t=10seconds.
 
Would I do the same then, for an amount like 25 or 35? Or do I say 25 would be 0, since there is no slope there, and then 35 would be a negative number, because the slope is going downward?
 
if t=25, all you need to do is look at the graph. what is happening at t=20? position is 100. now take t=25. position is still 100. what does that mean? in 5 seconds the position of the cyclist has not changed which means he is stationary. not moving so velocity is zero. The formula can also be applied. s=d/t. between 20 and 25 seconds. distance traveled is zero in a time frame of 5 seconds. so s=0/5 = 0. which is the same as saying rise over run. Between 20 and 30 would be the same, distance traveled between 20 and 30 seconds is zero in a time frame of 10 seconds so s=0/10=0. At t=35, the slope has changed. at t= 30 seconds position was 100 but now at t=35, the position is 50 again. So distance traveled is final distance minus the initial distance so 50-100=-50. And that has occurred in a time frame of 5 seconds. so again s=d/t, -50/5=-10 so velocity is -10.

That is why it has a negative gradient. So essentially, if the position at t=0 is 50 and then at t=25 is 100 and then at t=35 is 50 again it means he is going backwards. he has turned around at t=30 and is heading back
 
thanks a lot that help me a lot too. I fully understand it now
 

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