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Velocity vs time graph

  1. Sep 21, 2004 #1
    A graph can be located here:

    http://home.earthlink.net/~urban-xrisis/IMG_0373.jpg

    What is the average velocity between 0s and 50s?
    ALl I have to do is find the total distance travled and divide that by the total time passed correct?

    Also, the question asks "write an equation for x as a function of time for each phase of the motion, represented by (i) 0a, (ii) ab, (iii) bc"

    Any idea what it's trying to ask for? An equation for the derivative?
     
  2. jcsd
  3. Sep 21, 2004 #2
    Your first question: yes, total distance over time.
    Second question: an equation for x as a function of time is an equation for displacement with t as the variable; for example, x = 2t. For each section, find an equation of that form.
     
  4. Sep 21, 2004 #3
    Oops, sorry, average velocity is total displacement over time, not distance over time. A little different.
     
  5. Sep 21, 2004 #4
    so x is displacement and t represent time?

    why are variables needed? At the the specific time of point a, the time is 15s witha velocity of 50m/s. why is a formula needed?
     
  6. Sep 21, 2004 #5
    Yes, average velocity is total distance divided by total time. As for the other part if u look at the graph from 0 to a the velocity seems to be increasing linearly and the vel. is constant from a to b and from b to c is decreased linearly........so for...say... a to b for every t seconds x increases x units. Now for say 0 to a.....try to imagine the vel. getting greater and greater....what happens to the distance x gained for every t seconds??? Hint: What does the curve look like?.

    btw the first derivative of a x(t) function gives u v(t) and second derivative gives u a(t).......to go from V(t) to x(t) you need to integrate :)
     
  7. Sep 21, 2004 #6
    I dont know the formula of the graph so I cant find the derivative :( But I have the distance thing solved. For my second question, you said that "a to b for every t seconds x increases x units" do you mean that I should just find the slope?
     
  8. Sep 21, 2004 #7

    Pyrrhus

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    Do you know what the integration represents geometrically?

    what will the area under the curve be?
     
  9. Sep 21, 2004 #8
    distance travled? I still dont know what to do for the formula question
     
  10. Sep 21, 2004 #9
    Actually, total displacement over total time, since velocity is a vector value.

    For the equation part: you need an equation relating displacement and time. You have a graph of velocity-time. As said above, taking the integral of the velocity-time graph for each section will give you the area under the graph (although you can find that by simple graphical area analysis), which represents the displacement.

    Also, make sure you understand the difference between distance and displacement. The latter is a vector value, meaning it has direction as well as magnitude, whereas distance has only magnitude. This is important here because you have negative velocity.
     
  11. Sep 21, 2004 #10
    with the area of the graph, how do I come up with an equation for "write an equation for x as a function of time for each phase of the motion, represented by (i) 0a, (ii) ab, (iii) bc"
     
  12. Sep 21, 2004 #11
    I guess it is better to take the integral rather than graphically finding area. Find an equation for the v-t graph for each interval, then take the integral of that to get a d-t equation. Jeez, I should have been more clear from the beginning. I didn't think the problem through correctly.
     
  13. Sep 21, 2004 #12
    2 problems. I am not given the equation of the graph, and I dont know how to to integral (unless that's the same thing as taking the derivative)
     
  14. Sep 21, 2004 #13

    Pyrrhus

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    If you haven't been taught how to integral, then maybe consider two triangles and rectangle and see if you can relate Time and V with them, considering Displacement or think it as X.
     
  15. Sep 21, 2004 #14
    that's what I did at the end but what about finding the formula, I still have no clue what it wants. How does finding the displacement from zero to point a give me an equation of how position relates to velocity?
     
  16. Sep 22, 2004 #15
    You need an equation to relate displacement and time, not position and velocity. Gotta go, I'll come back later.
     
  17. Sep 22, 2004 #16
    This gets a little complicated if you can't take the integral. You can find the area under the v-t graph for the first section (for example), but since the velocity is not constant, your d-t graph will not be a straight line, but a curve. Therefore, you can find the area traveled in 15 s, place that point on your d-t graph, then estimate a curve for d-t up to that point. The constant velocity part will be easier, as your d-t graph will be linear, but you need to estimate the curve for the last interval again. I'm not sure if you can really develop an equation for those sections based on that.
     
  18. Sep 22, 2004 #17
    Yeah, first, you gotta find an equation relating velocity and time for each of the 3 intervals. Then you gotta integrate each one to find the relationship between displacement and time. Do you know how to integrate?
     
  19. Sep 22, 2004 #18
    He needs to do it by graphical analysis.
     
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