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Force/time graph acceleration problem

  1. May 25, 2017 #1
    1. The problem statement, all variables and given/known data

    Figure 5-55 gives, as a function of time t, the force component
    Fx that acts on a 3.00 kg ice block that can move only along
    the x axis. At t = 0, the block is moving in the positive direction of
    the axis, with a speed of 3.0 m/s.What are its (a) speed and (b) direction
    of travel at t = 11 s?

    problem63.png

    2. Relevant equations

    ##\vec{F}=m \vec{a}##
    ##a_{avg}= \frac{a_1-a_0}{t_1-t_0}##
    ##V = V_0 + \vec{a}t##

    3. The attempt at a solution

    So, if I understand the problem correctly the graph is giving me Fx as a funcion of t, assuming that Fx is the only horizontal force acting on the block a should be given by:

    ##\vec{F}=m \vec{a}## → ##\vec{a}= \frac {\vec{F}}{m}##

    And I think is safe to use the slope formula for the non-horizontal changes in a because they are just a line at a slope, so aavg should be equal to a
    My thought process for the problem is that there is not a "smart way" to solve it, we have to just evaluate V from start to finish in order to get V11

    So, starting with the problem:

    V0 = 3 m/s
    m = 3 Kg

    a0 = 2/3 m/s2
    a2 = 2 m/s2
    aavg0/2 = 2/3 m/s2
    a2/5 = 2 m/s
    aavg5/6 = -2 m/s2
    aavg6/9 = -4/3 m/s2
    aavg9/11 = 4/3 m/s2

    Now that we have the change in acceleration for each part of the graph we can simply calculate the final velocity (I'm using unnecessary parenthesis in the formula just to make more clear the different time frames):

    V11 = 3 m/s + (2/3 m/s2 ⋅ 2s) + (2 m/s ⋅ 3s) + (-2 m/s2 ⋅ 1s) + (-4/3 m/s2 ⋅ 3s) + (4/3 m/s2 ⋅ 2s) = 7 m/s

    So the answer should be
    (a) s11 = 7 m/s
    (b) Positive direction


    But the answer in the book (Halliday,Resnik, Walker 9th Ed) is 8 m/s, and I really have no idea what could be wrong in my thought process and calculation, any clue?
     
  2. jcsd
  3. May 25, 2017 #2

    CWatters

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    Have you studied the impulse-momentum theorem?
     
  4. May 25, 2017 #3
    No, if it helps the problem is in the "Force and motion 1" section, and I assume you are supposed to learn what you suggested later on
     
  5. May 25, 2017 #4

    CWatters

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  6. May 25, 2017 #5
    I could do it this way and I will likely try to do so later, but I'm pretty sure the problem is intended to be solvable without that information, I'd assume you learn about that theorem in the kinetic energy section, which comes later on.

    Anyway, do you have any idea about solving this problem without that theorem?
    I would like to understand the thought process behind solving it with just the knoledge I'm supposed to have at this point more than just solving it easily because it becomes trivial once you have more informations
     
  7. May 25, 2017 #6

    CWatters

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  8. May 25, 2017 #7
    By using calculus it's easy, we know that the rate of change of acceleration is constant (from shape) => let da/dt=j => a=jt (no initial rate of change of acceleration) integrate with respect to time => v=1/2jt2 but a=jt => j=a/t so v=1/2at. But I don't think he is supposed to do this way
    With initial rate of change of acceleration v=j0+1/2at
     
  9. May 25, 2017 #8
    By the way, I didn't had time to post this earlier but I understood the problem in my logic, of course it's wrong to use the average formula, because the acceleration is not constant, in this case it's hard for me to find a solution that avoids calculus
     
  10. May 25, 2017 #9

    haruspex

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    Yes, it is an integral (of acceleration wrt time, or, equivalently, of force wrt time), but because it's all straight lines you do not need any calculus algebra to do the integrals. Just take the areas.
     
  11. May 26, 2017 #10

    jbriggs444

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    Pretty sure you can.
     
  12. May 26, 2017 #11

    CWatters

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    Oops you're correct. I've checked and it gives the right answer.
     
  13. May 26, 2017 #12
    So my thought process was actually correct but I made some misteps in the calculations?

    If yes, I can't really come up with the algebraic mistakes I made, and at the moment I am not even sure why my original thoughs are actually ok
     
  14. May 26, 2017 #13

    CWatters

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    OK so..

    The average acceleration in phase 1 isn't 2/3m/s2. It starts at 2/3m/s2 and ends at 6/3m/s2 so the average is 4/3m/s2

    There are also 6 phases as the slope between 5 and 6 isn't the same as 6 and 7 seconds.

    The acceleration between 5 and 6 is +ve. The force maybe reducing but it's still +ve until 6 seconds.
     
  15. May 26, 2017 #14

    jbriggs444

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    What is the average of 2/3 and 2?

    Huh? That acceleration is positive and the average is not two. Did you switch from calculating averages to calculating slopes?
    Huh? There is a bend in the acceleration graph between 6 and 9.
    The acceleration over this part of the graph is negative, not positive and the average is not 4/3.
    You have the average rate of change in velocity for each part of the graph. Which has to be multiplied by time. [The equation shows that you knew that. You just didn't say it right]
     
  16. May 26, 2017 #15
    I have some questions

    You say that a in phase 1 is 4/3m/s2, may I ask why?
    The Δa is 4/3m/s2, but it should be considered during the Δt = 2s if I undestand it correctly, so the total Δv during the 2 seconds is 4/3m/s2,
    In fact in my calculations I am adding 4/3m/s2 in total for phase 1.
    Are you telling me that the Δv is instead twice as much?

    Then, I am considering the slope between 5 and 6 in this part of the equation
    (-2 m/s2 ⋅ 1s)

    But then I am combining 6/7 with 7/9 in this part of the equation
    (-4/3 m/s2 ⋅ 3s)
    I skipped some steps, but the reasoning behind it was that in the 6/7 phase a is
    ##\frac{\frac{-4}{3}-0}{1s} = \frac{-4}{3}##
    And in the phase 7/9 a is just -4/3, so it was reasonable for me to combine the two phases in one step

    And for the last part, I'm not sure if I'm following you, but in the phase 5/6, despite the force being positive, it should decrease the acceleration because less force is being applied right?
     
  17. May 26, 2017 #16
    I'll do it step by step, I've always used this formula

    ##a_{avg}=\frac{a_1−a_0}{t_1−t_0}##
    So, in the first case I get
    ##a_{avg}=\frac{2−2/3}{2−0}=2/3##

    I guess this will answer all of your questions, am I calculating aavg wrongly?
     
  18. May 26, 2017 #17

    haruspex

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    Time to stop, then. Did you really mean that?
     
  19. May 26, 2017 #18
    Well, I want to punch myself for not using Δv now that i think about it.

    My god what a stupid mistake...

    So, let's say that the problem is just solved if I just use
    ##a_{avg}= \frac{v_1-v_0}{t_1-t_0}##
     
  20. May 26, 2017 #19

    jbriggs444

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    The starting and ending velocities are what you are trying to find. What you are given are the starting and ending accelerations. For an acceleration with a constant rate of change (a straight line on an acceleration-time graph):

    [Corrected, thanks @haruspex]
    $$a_{avg}= \frac{a_1+a_0}{2}$$
    You use final minus initial if you are looking for a rate of change.
    You use final plus initial if you are looking for an average.
     
    Last edited: May 26, 2017
  21. May 26, 2017 #20
    Oh, I see.
    I tend to get tricked by problems involving average values because I actually get to practice it much less than other problems, and this time I made the wrong assumption that my math was right.


    Thanks a lot for the help!
     
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