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Velocity vs Time

  1. Feb 20, 2017 #1
    Hey guys I have attempted to solve the question below and hoping to get some feedback. Also, any pointers to solve question 'c' would be really appreciated.

    1. The problem statement, all variables and given/known data

    1. A sports car's velocity is shown in the graph below. (Please see the image)


    a. How fast is it traveling at the 5 s mark?
    b. What is the acceleration at the 5 s mark?
    c. Estimate the acceleration at the 15 s mark. Show your work.
    d. During which intervals of time, did the car have a negative acceleration?​

    2. Relevant equations
    Acceleration = Δv/Δt

    3. The attempt at a solution

    a. 4.5 s
    b. 0.9 m/s2
    c. I cant seem to understand this. Not sure what is the initial velocity in this case. Any help would be appreciated.
    d. 11, 12, 13, 19, 20
  2. jcsd
  3. Feb 20, 2017 #2


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    For c) what is acceleration?

    For d) your answers are not time intervals.
  4. Feb 20, 2017 #3
    Acceleration will be the slope of the line at any point.

    a. Are you sure about 4.5?
    b. There appears to be a slight curve in the plot from v=0 to v=5. But it is very slight. So if I assume it is straight from t=0, v=0 to t=7.2, v=5, then I get an acceleration of about 0.7. I'm not sure how you got something as high as 0.9.
    c. Same thing as step b. Pick 2 points around the 15s mark that appear to be a straight line that closely represent the slope of the curve at the 15s mark and find the slope of that line.
    d. It is asking for time intervals where the acceleration is negative. So, for example, you may say that from t=1 to t=4 the car has a negative acceleration (although what I just wrote is not true.)

    Edit: I think I just realized how you got 0.9. I think you got it from the 4.5 answer in part a.
  5. Feb 20, 2017 #4


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    Check this again. Also, the unit should be m/s.
    This changes too, because of a).
  6. Feb 20, 2017 #5
    Thanks a lot guys. I double checked my work and here are my answers:

    a. 3.5 m/s
    b. 3.5/5 = 0.7 m/ss
    c. I picked two points: 7.25 - 5.25/15-14 = 2 m/s2
    d. t=20.5. t=11, t=12, t=18.5, t=19, t=20
  7. Feb 20, 2017 #6


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    If this is a calculus based course what I would kind of do is this...

    Now what you have is a series of straight lines instead of the smooth curves. You should now be able to find the equations for those lines, (i.e. to about 9 seconds the equation is just roughly ##v=\frac{2}{3}t##) take the derivative to find the acceleration. For example, at 15 seconds, you have a specific curve that should be easy to find the equation of using any of the techniques in algebra. Take the derivative and you'll find the acceleration for any time on that curve. Your approximate lines should be accurate enough for an estimation.

    Alternatively, it might help to plot the acceleration graph on the same graph.

    Do this make sense? I know I didn't draw it very well....

    Whoops, others already replied..
  8. Feb 20, 2017 #7


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    Did you see PeroK's comment on D?
  9. Feb 20, 2017 #8
    Oh I see. So it will be t = 10.5 to t=12 and t=18.5 to t=20?
  10. Feb 20, 2017 #9


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    Looks about right.
  11. Feb 20, 2017 #10
    what about my answers for a, b, and c..are they correct?
  12. Feb 20, 2017 #11


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    Looks good enough to me.
  13. Feb 20, 2017 #12
    If it was me, I would say 10.5 to 13 and 18 to 20. The decrease from 12 to 13 is pretty obvious.
  14. Feb 20, 2017 #13
    Thanks a lot everyone for the feedback.
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