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Velocity vs. time graph questions

  1. Oct 2, 2007 #1
    [​IMG]

    A. Q: FIND THE maximum velocity. During what time interval does this velocity occur?
    A: The maximum velocity is 16 m/s, and it occurs between 35 and 50 seconds.
    Am I right?

    b. Q: The velocity is negative from 30 t0 35 s. What does this mean?
    A: I think this means that the car is backing up, and goes past it's reference point, from where it started.

    c. Q: Find the displacement during the following time intervals:
    A: 0s to 14s = v=d/t so d= vt = (9m/s)(14s) = 126 m
    14s to 30s = d=vt = (0m/s)(16s) = 0m
    30s to 35s = d =vt = (-4m/s)(5s) = -20m
    35 to 50s = d=vt = (16 m/s)(15s) = 240m


    d. Q: What is the total displacement?
    A: 126m+0m+(-20m)+240m = 346m

    e. Q: What is the average velocity?
    A: 9m/s + 0m/s + -4m/s + 16 m/s = 21 m/s / 4 = 5.25 m/s
    Is this the right way to find the average velocity? Or should I find it with another method?
     
  2. jcsd
  3. Oct 2, 2007 #2

    ptr

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    Average velocity is total displacement by total time. The others look correct.
     
  4. Oct 2, 2007 #3

    Kurdt

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    Staff Emeritus
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    Gold Member

    Yes that looks right.

    I'm not sure what you mean by reference point, all you would need is to know its moving in the opposite direction.

    These are correct.

    Correct again.

    Average velocity is:

    [tex]v_{ave} = \Delta x / \Delta t[/tex]
     
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