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Homework Help: How to graph velocity where there is a change in acceleration

  1. Sep 16, 2014 #1
    The speed is a constant of 20m/s for 100m (moving on a straight line) after 100m, the car stops at a constant 6m/s/s.
    *So in this case, speed is equal to velocity?
    *How do you graph velocity vs time graph and acceleration vs time graph?
  2. jcsd
  3. Sep 16, 2014 #2
    The problem statement tells you specifically how the acceleration is varying with time. What is your quantitative understanding of this?
  4. Sep 16, 2014 #3
    But first I need to know if constant speed is equal to velocity. So if there is a constant speed of 20m/s , is it mean that the velocity is constant 20m/s ?
  5. Sep 16, 2014 #4
    The problem implies that this is the case.

  6. Sep 17, 2014 #5
    So for the displacement vs time graph, the line would be going up from 0 to 100 ( take 5 sec) and then from 6 sec, it starts going down by 6 because of the acceleration. 5s=100m, 6s=94m and so on?
  7. Sep 17, 2014 #6
    From 0 to 5 sec., this is correct. After five seconds, it is incorrect.

    What equations are you using? If you had followed the PF template, you would have written down your equations in advance, and it would have been easier to help you. Please write down your equations.

  8. Sep 17, 2014 #7
    The second one? Df=.......

    Attached Files:

  9. Sep 17, 2014 #8
    Yes, but you would have to restart the application of the equation after the first 5 seconds, so d in the equation would be the distance relative to the location after 5 seconds, and t in the equation would be 0 at the running time of 5 seconds.

    Hope this makes sense. Now, what do you get?

  10. Sep 17, 2014 #9
    So we r starting new after 5sec? Df=100 ,vi=20, tf=0 ,a= -6 substitute these numbers?
  11. Sep 17, 2014 #10
    Wait no,,, di =100 , df is unknown
  12. Sep 17, 2014 #11
    ..wait ..so since im starting another one,, everything before 5sec doesnt matter,, so my vi is not 20m/s anymore?
  13. Sep 17, 2014 #12
    You seem to be pretty confused. But, you are beginning to get the right idea.

    I might be better to think of this as two separate problems:

    Problem 1:
    Applies between time 0 and 5 seconds
    Initial distance = 0
    Velocity = 20 meters/sec
    Acceleration = 0

    You have already solved problem 1 correctly

    Problem 2
    Applies after 5 seconds
    Initial distance = 100
    Initial velocity = 20
    Acceleration = -6

    You need to modify your equations a little to handle this part:

    v = 20 - 6t'
    d = 100 + 20t'-6(t')2/2
    where t'=(t-5)


    v = 20 - 6(t-5)
    d = 100 + 20(t-5) - 6(t-5)2/2

    Do these equations make sense to you?

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