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Fundamental kinematics question

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    I seem to have a fundamental misunderstanding of the kinematic principles in this question.

    A constant force of 8.0N is exerted for 4.0s on a 16-kg object initially at rest. What will the change in speed of this object be?

    2. Relevant equations
    F = ma
    Δx = v0t + 1/2 at2
    Δv = v0 +at
    Δv = Δx/Δt
    Δv = v0 + aΔt

    3. The attempt at a solution
    F = ma

    thus

    a = (8.0N) / (16.0 kg) = .5 m/s^2

    then

    Δx = (0 m/s)(4 s) + 1/2(.5 m/s^2)(4 s)^2 = 4 m

    if

    Δv = Δx/Δt

    then

    Δv = (4 m) / (4 s) = 1 m/s

    but using Δv = aΔt

    Δv = (.5 m/s^2)(4 s) = 2 m/s

    I don't understand why I would have two conflicting answers there. Just curious if anyone might have some insight on why that would be. Thanks!
     
  2. jcsd
  3. Dec 10, 2014 #2

    Stephen Tashi

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    [itex] \frac{ \triangle x}{\triangle t} [/itex] is an average velocity over the 4 second interval, not the instanteous velocity of object at the end of the 4 seconds.
     
  4. Dec 10, 2014 #3

    Bystander

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    What's this equation mean to you?
     
  5. Dec 10, 2014 #4

    Stephen Tashi

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    There are various interpretations. One is that if velocity is constant with respect to time then it is equal to the change of distance over a time interval divided by the length of that time interval. Thinking of the equation as a way to remember some calculus, you could think of the meaning: "Instanteous velocity at a given time is equal to instantaneous rate of change of distance with respect to time at that instant". The way you used the equation was to compute an average velocity.
     
  6. Dec 10, 2014 #5

    Bystander

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    I was asking the OP, not you, as a hint, while we posted simultaneously.
     
  7. Dec 11, 2014 #6

    Stephen Tashi

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    I should have noticed that you weren't the OP.
     
  8. Dec 11, 2014 #7

    ehild

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    Your relevant equations are not correct. It is uniformly accelerating motion, that means v = v0 +at, where v0 is the velocity at t=0, or change of velocity in the interval Δt is Δv=aΔt.
    The displacement Δx= v0t + 1/2 at2 is correct.
     
  9. Dec 11, 2014 #8
    Thanks for the quick reply. I suspected that at first as well. I then thought that if 2 m/s was the final velocity, then

    Δv = (2 m/s) - (0 m/s) = 2 m/s

    I just couldn't see how 1 m/s could be the average, but I'll play around with it some more and see how that happens. Thanks again.
     
  10. Dec 11, 2014 #9
    For v = v0 + at, that would play out as:

    v = (0 m/s) + (.5 m/s2)(4 s) = 2 m/s

    I'm confusing average velocity with change in velocity. I new there was something. Thanks again folks!
     
  11. Dec 11, 2014 #10
    For the average of two velocities, it would be something like:

    vavg = (2 m/s - 0 m/s)/2 = 1 m/s
     
  12. Dec 11, 2014 #11

    ehild

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    The change of speed was asked.
    Yes, the speed will be 2 m/s at the end of the 4th second. As it was zero initially, so the change of speed is 2 m/s.

    The average velocity is defined for a time interval Δt as displacement Δx over time Δt: vaverage=Δx/ Δt.
    You have the formula for Δx=voΔt+a/2 Δt2. As vo = 0, Δx=a/2 Δt2, and vaverage=Δx/ Δt=a/2 Δt.
    There is a other formula for the displacement in case the velocity changes from v1 to v2 in time Δt: ##\Delta x = \frac{v_1+v_2}{2}\Delta t##.
    In case of uniformly accelerating motion, the average of the velocity in a time interval is also the average (mean) of the initial and final velocities.
     
  13. Dec 11, 2014 #12
    Roger that. Good to know!
     
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