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Homework Help: Kinematics motion

  1. Dec 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider an object traversing a distance L, part of the way at speed v1 and the rest of the way at speed v2. Find expressions for the average speeds when the object moves at eac of the two speeds.

    a) for half the total time
    b) for half the distance

    2. Relevant equations


    3. The attempt at a solution

    xf - i = [(vi + vf)/2]t
    L = [(v1 + v2)/2]t
    L = [(v1+v2)/2]0.5t

    2L/t = [v1 + v2]/2

    L/2t = [v1 + v2]/2
  2. jcsd
  3. Dec 16, 2013 #2
    Responding on my phone while watching a movie, but I believe your equation doesn't apply to this situation, and that's what's wrong.

    That equation is an expression of x=vt using an average for v, final minus initial divided by two. That only applies under a constant acceleration, for which you do not har, you accelerate at one point on the trip, from initial velocity to final velocity instantaneously.

    Instead, try using a simple x=vt but treat each distance/time you travel at the initial or final velocity as one set of constant acceptation equations.

    You know they have the same amount of time for one of the problems, and the same amount if distance for the other problem.
  4. Dec 16, 2013 #3
    This would mean that object is traveling half of the time at speed v1 and other half at speed v2?
  5. Dec 16, 2013 #4
    Yeah that's the key. The equation will work only for the time part, but half and half of the distance will not work user the same assumption.
  6. Dec 16, 2013 #5


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    Right so far.
    Why the extra 0.5?
    That (apart from an extra 2 that's magically appeared) is the 'half the time at each' equation again.
    Let the total distance travelled be d. How far does the object go at v1? How long will that take?
  7. Dec 16, 2013 #6
    Because part(a) wanted half of total time?
    Hang on, isn't L the total distance?
    Last edited: Dec 16, 2013
  8. Dec 16, 2013 #7
    I'm lost. Could someone provide an exposition of this in greater clarity?
  9. Dec 16, 2013 #8
    The average between two velocities is weighted by the time the car speds at that velocity, not the distance traveled at that velocity.

    So if the distance for both velocities is the same, half the total distance for each, then the car must spend less time going fast then slow, because it would traverse that distance quicker in the fast velocity.

    So you gotta use a different equation, not the verge velocity one!
  10. Dec 16, 2013 #9
    But "...two velocities is weighted by the time the car speds at that velocity" is by definition, at least mathematically, v.t which produces displacement.

    But let's see:

    first leg:

    dx1 = [xf - xi]1 = v1t + 0.5(0)t^2
    dx1 = v1t

    second leg:
    dx2 = [xf - xi]2 = v2t + 0.5(0)t^2
    dx2 = v2t


    L = v1t + v2t

    average velocity = total displacement /total time

    [v1t + v2t]/t but question asked for average velocity for half the total time

    [v1t + v2t]/(t/2)

    average velocity = 2(v1 + v2)

    Am I right up till here?
    Last edited: Dec 16, 2013
  11. Dec 16, 2013 #10
    If you're working with the same time for both, yes! But if both legs have the same amount of time spent at them, you can just use the average velocity equation you had earlier.

    If they go the same distance, they must have differing times, so you'll be a t_1 and t_2, but then can set them equal to each other as well.
  12. Dec 16, 2013 #11
    xf - i = [(vi + vf)/2]t
    L = [(v1 + v2)/2]t

    (v1 + v2)/2 = L/t

  13. Dec 16, 2013 #12
    Yes, but only for the part of the problem where the car spends equal time at each velocity.
  14. Dec 16, 2013 #13
    Isn't this what part (a) is asking?
  15. Dec 16, 2013 #14
    Yes, it sure is. But part b has tone tackled a different way, setting the distances equal instead.

    I'm sleeping now, but ill be back in the morning hopefully :)
  16. Dec 16, 2013 #15
    I've to leave the house now after being coerced into some social function. I'll be working on part (b) on the train.

    But if the answer to part (a) is (v1 + v2)/2 = L/t, why does the answer sheet states v = [v1 + v2]/2?
  17. Dec 16, 2013 #16

    First leg of the journey:
    L/2 = v1t
    t = L/2v1
    (remaining time for second leg= 2v1-L)
    L/2 = v1[L/2v1]

    Second leg of the journey:
    L/2 = v2[2v1-L]

    Average velocity = total displacement/ total time=
    [v1(L/2v1)+ v2(2v1-L)]/[2v1]
    Last edited: Dec 16, 2013
  18. Dec 16, 2013 #17
    Because L/t is v. Average velocity over a distance is always going to be the length traveled over the time it took.
  19. Dec 16, 2013 #18
    Could you show me ho you got the time for t2? Also, double check your dimension on that equation.
  20. Dec 16, 2013 #19


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    Since the times will be different, better to write t1 here.
    v1 is a speed, L is a distance. How can you get a time, or anything meaningful, by subtracting a distance from a speed?
    Calculate the time for the second leg, t2, the same way you calculated t1.
  21. Dec 16, 2013 #20
    You're right. It's tautological.
  22. Dec 16, 2013 #21
    second leg of the journey:

    L/2 = v2t
    t2 = L/2v2
    L/2 = [v2L]/2v2

    Average velocity = displacement / total time

    L/2 + L/2 = [v1L]/2v1 + [v2L]/2v2 = [2v2(v1L) + 2v1(v2L)] / 2(v1v2)

    total time = t1 + t2 = (L/2v1) + (L/2v2)


    [ [2v2(v1L) + 2v1(v2L)] / 2(v1v2) ] / [ (L/2v1) + (L/2v2) ]

    reducing it gives me [2v1v2] / v1 + v2

    Last edited: Dec 16, 2013
  23. Dec 16, 2013 #22
    t2 = L/2v2

    it's solved
  24. Dec 16, 2013 #23


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    Bingo. In case you don't know, that expression is the 'harmonic mean' of v1 and v2.
  25. Dec 17, 2013 #24
    It's really just statistic isn't it?
    I've never touched stats and don't quite intend to. It's messy, chaotic, and inelegant. I would prefer the more abstract beauty of pure mathematics language.
  26. Dec 17, 2013 #25


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    I'm not aware of any connection with statistics.
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