Instantaneous velocity and average velocity

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Homework Help Overview

The discussion revolves around determining instantaneous and average velocities from a graph depicting the position of a rabbit over time. Participants are exploring how to interpret graphical data to find these velocities without relying on equations.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the concept of instantaneous velocity as the slope of the tangent line at specific points and average velocity as the slope between two points. There are questions about how to accurately determine these slopes graphically, particularly when the graph does not provide explicit equations.

Discussion Status

Some participants have offered guidance on estimating slopes from the tangent lines drawn on the graph. There is a recognition of the challenge in achieving precise values, as one participant expresses concern over discrepancies in their calculations compared to expected answers.

Contextual Notes

Participants note that the problem requires a graphical approach to determine instantaneous velocity, which may lead to variations in results due to estimation methods. There is an emphasis on the need for accuracy in the context of homework expectations.

Struggling
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hi all having a problem with this question:

The position of a rabbit along a straight tunnel as a function of time is plotted: http://img284.imageshack.us/img284/5764/untitled0vz.png
1. What is its instantaneous velocity at t = 10.0s and t=30.0s
2. what is the average velocity:
between t = 0 and t = 5.0s
between t = 25.0s and t = 30.0s
between t = 40.0s and t = 50.0s

i can do it when they give an equation. and i know it has to all do with tangents and all that, i can get the results by "estimating", but not as accurate. I am totally stuck with this one any help at all will be appreciated.

thanks
 
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Instantaneous means the derivative at that point in time; average is the slope formed between the points.
 
Knavish said:
Instantaneous means the derivative at that point in time; average is the slope formed between the points.

yeh i know that but how do you get the equation? i understand taking 2 points eg. (x1,y1)(x2,y2) but i can't find a reliable way of finding them
 
Struggling said:
yeh i know that but how do you get the equation? i understand taking 2 points eg. (x1,y1)(x2,y2) but i can't find a reliable way of finding them
this problem is to determine instant velocity GRAPHICALLY.
you don't need any "equations" in the usual sense.
instant velocity is the SLOPE of the tangent line at a point along the curve.
just determine the slope of the tangent line using (Δx/Δt) by choosing 2 points along the tangent line (1 point can be that at which the tangent line is tangent to the curve) and calculate (Δx/Δt)=(x2 - x1)/(t2 - t1). see this diagram:
http://www.andamooka.org/newtphys/figs/bk1/ch02/motiond.JPG
 
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Apparently the tangent lines are already drawn on the graph at t= 10 and 30. For the tangent line at t= 30, it appears to me that the line crosses y= 0 around x= 18 and crosses y= 25 around x= 36. The slope of the tangent line is 25/(36-18)= 25/18 or about 1.39 and that is the instantaneous speed at t= 30. That's about the best you can do given that information.
 
HallsofIvy said:
Apparently the tangent lines are already drawn on the graph at t= 10 and 30. For the tangent line at t= 30, it appears to me that the line crosses y= 0 around x= 18 and crosses y= 25 around x= 36. The slope of the tangent line is 25/(36-18)= 25/18 or about 1.39 and that is the instantaneous speed at t= 30. That's about the best you can do given that information.

you serious? that's what i did exactley but they have answers to 2 decimal places and i kept getting 0.01-0.10 off the answer and i was stressing sooo much i was 90% certain that there must have been a formula.
 

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