Instantaneous velocity and average velocity

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SUMMARY

The discussion focuses on calculating instantaneous and average velocity from a graphical representation of a rabbit's position over time. Instantaneous velocity is determined by the slope of the tangent line at specific points, specifically at t = 10.0s and t = 30.0s. Average velocity is calculated between defined time intervals: from t = 0 to t = 5.0s, t = 25.0s to t = 30.0s, and t = 40.0s to t = 50.0s. The participants emphasize that no traditional equations are necessary, as the graphical method suffices for accurate results.

PREREQUISITES
  • Understanding of graphical analysis in physics
  • Knowledge of derivatives and slopes
  • Familiarity with the concept of average velocity
  • Ability to interpret position-time graphs
NEXT STEPS
  • Study the concept of derivatives in calculus for better understanding of instantaneous velocity
  • Learn how to calculate slopes of tangent lines on graphs
  • Explore graphical methods for determining average velocity over intervals
  • Review examples of position-time graphs in physics to reinforce concepts
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Students studying physics, particularly those focusing on kinematics, educators teaching motion concepts, and anyone interested in graphical analysis of velocity.

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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