Instantaneous velocity and average velocity

  • Thread starter Struggling
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  • #1
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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 [Broken]
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. im totally stuck with this one any help at all will be appreciated.

thanks
 
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  • #2
Chronos
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  • #3
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Instantaneous means the derivative at that point in time; average is the slope formed between the points.
 
  • #4
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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 cant find a reliable way of finding them
 
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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 cant 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 [Broken]
 
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  • #6
HallsofIvy
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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.
 
  • #7
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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? thats 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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