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Acceleration velocity vs. time graph

  1. Oct 4, 2007 #1

    a. Q: Find the acceleration between 0s and 4s.
    A: vf= 10m/s vi=0m/s
    a = vf - vi / t
    = (10m/s - 0m/s) / 4s
    = 10 m/s / 4 s
    = 2.5 m/s^2

    b. Q: Find the acceleration at 8s.
    A: a = v/t
    = (6.5m/s)(8s)
    = 52 m/s
    = 50 m/s^2 (significant digits?)
    PS. The line is going down, but the velocity is still in the postive x-axis. How do I represent that it is going in the opposite direction?

    c. Q: Find the acceleration at 5s
    A: a = v/t
    = (10m/s)(5s)
    = 50 m/s^2

    d. Q: Find the average velocity for this trip. (Hint: First find the total displacement)
    A: Do I have to find the displacement for each section of the graph?, and add it together for total displacement, then divide it by 20s (total time), to find the average velocity?
  2. jcsd
  3. Oct 4, 2007 #2


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    Staff Emeritus
    Science Advisor


    The acceleration is negative.


    The total displacement is the area under the graph. Yes, find the area of each individual section. You can do that in only two parts, using the formula for the area of a trapezoid. Subtract the area below the t-axis, don't add it! Finally, divide by 20s.
  4. Oct 4, 2007 #3


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    Homework Helper

    For parts b and c, why are you multiplying velocity by time instead of dividing?

    remember that acceleration is the slope of the v-t graph... you did part a) correctly... do parts b) and c) the same way.
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