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A Truck Braking

  1. Jan 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A truck is initially moving at velocity v. The driver presses the brake in order to slow the truck to a stop. The
    brake applies a constant force F to the truck. The truck rolls a distance x before coming to a stop, and the time
    it takes to stop is t.
    8. Which of the following expressions is equal the initial kinetic energy of the truck (i.e. the kinetic energy before the driver starts braking)?
    (A) F x
    (B) F vt
    (C) F xt
    (D) F t
    (E) Both (a) and (b) are correct
    9. Which of the following expressions is equal the initial momentum of the truck (i.e. the momentum before the driver starts braking)?
    (A) F x
    (B) F t/2
    (C) F xt
    (D) 2F t
    (E) 2F x/v

    2. Relevant equations
    W = delta K
    p= m*v
    F*t = delta p
    3. The attempt at a solution
    For 8 I used the work energy theorem to get the kinetic energy to equal Fx. And since x=vt, I picked (E) but apparently (A) is right. Why doesn't x=vt apply?
    For 9 I used the fact that force* time = delta p, to get that p=F*t, and that's where I get stuck. Any thoughts on how to continue?
     
  2. jcsd
  3. Jan 18, 2015 #2
    The final velocity of the truck is 0 . So, the final momentum is also 0 . And ##\Delta p = p_{final} - p_{initial} = 0 - p_{initial} = - p_{initial}## . Here, the sign is negative because the force is applied opposite to the direction of the initial velocity of the truck.
    And look, ##x = (\frac{v_{initial} + v_{final}}{2}) \cdot t ##
     
    Last edited: Jan 18, 2015
  4. Jan 18, 2015 #3

    Nathanael

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    The velocity is not constant.

    Which answers can you eliminate?
     
  5. Jan 18, 2015 #4
    Oh ok I get 8 then, basically x=vt doesn't apply when accelerating. OK.
    As for 9 I can eliminate answers A and C from dimensional analysis.
     
  6. Jan 19, 2015 #5

    Nathanael

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    8.(B) would be correct if v was the average velocity instead of the initial velocity.

    Ok good. You said you think p=F*t, so 9.(B) and 9.(D) don't agree.
    What about 9.(E), that seems like a strange one, what are your thoughts on the meaning of that answer?
     
  7. Jan 19, 2015 #6
    I would have said that they're using t=x/v but since we have established that only works for average velocity, I don't really know.
     
  8. Jan 19, 2015 #7

    Nathanael

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    Right, but we know that the force is constant. Can you use that to determine what the average velocity is?
     
  9. Jan 19, 2015 #8
    Since there is constant acceleration due to a constant force and the final velocity is 0, isn't the average v just equal to v_initial/2?
     
  10. Jan 19, 2015 #9

    Nathanael

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    Good. So you said t=x/vavg and you said vavg=vi/2, then what is x/vi?
     
  11. Jan 19, 2015 #10
    It is equal to t/2.
     
  12. Jan 19, 2015 #11

    Nathanael

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    And so 9.(E) is....?

    (So many problems at once! :) )
     
  13. Jan 19, 2015 #12
    Since p=Ft and x/v_i=t/2 then t equals 2x/v_i and the answer is just p=2x/v_i. But doesn't (E) refer to the initial velocity and not the average velocity?
     
  14. Jan 19, 2015 #13

    Nathanael

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    I think you mean p=F*(2x/vi) ?

    Right. Aren't we referring to the initial velocity when we say vi?
     
  15. Jan 19, 2015 #14
    Ok yes to both counts, just out of curiosity if Ft was an answer choice would that be right?
     
  16. Jan 19, 2015 #15

    Nathanael

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    Yes Ft would also be right, but I guess they didn't want to make it too easy :)
    (At least they didn't give an option of "none of the above," that would probably trick a number of students!)

    2Fvi/x is just a strange way of writing Ft
     
  17. Jan 19, 2015 #16
    Ok well the Physics Olympiad (where I got the questions) loves to make things difficult. Anyways, thanks for your help!
     
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