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Questions in linear motion and rotating motion

  1. May 24, 2012 #1
    Hi

    please I want from you check my answer

    http://store2.up-00.com/May12/Ueu86896.jpg [Broken]

    http://store2.up-00.com/May12/fkl86896.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 24, 2012 #2
    In this case the equation
    V2=u2+2as
    means that the square of the final velocity equal the square of the initial velocity plus two times the stopping acceleration time the distance over which the car stopped. Think through the question and see that the car does not accelerate over a distance of 60 m.

    If you substitute the correct distance, you'll get the correct answer.
     
  4. May 25, 2012 #3
    But how I can know the total distance where the car stop
     
  5. May 25, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, you do. You are told that the driver sees a person 60 m ahead of him. You are also told the he continues at 20m/s for 1.5 s before braking. At 20 m/s he will go (20)(1.5)= 30 m in 1.5 seconds. He has the remaining 60- 30= 30 m in which to stop.
     
  6. May 25, 2012 #5
    You are using the same variables, s and t, for given values and required values.
    You should consider what is meant by delay.
     
  7. May 25, 2012 #6
    1 )

    u = 20
    V = 0
    s = 30 m

    a ) V^2 = u^2 + 2as
    0 = 20^2 + 2 X a X 30
    a = -6.6

    time requried

    V = u + at
    t = 3 s

    now distance

    s = ut + 1/2 a t^2
    s = 20 X(3) + 1/2 X(-6.6) X 3^2
    s = 30 m

    uniform accleration = a = -6.6
     
  8. May 25, 2012 #7
    Correct except the answer rounds to -6.7 m/s2
    Correct.

    Yes, but you have already calculated ( or, rather, HallsofIvy calculated) how far the car takes to stop. You did not need to do it again.
     
  9. May 25, 2012 #8
    thank sou much what about second one ...
     
  10. May 25, 2012 #9
    I am wrestling with the wording a little...You have correctly calculated the average rotational velocity as 20 rad/s for the first 10 seconds. The given information says the next eight seconds has a lower average rotational speed. In other words, there was a negative acceleration, so your answer that angular acceleration is zero is not correct.

    What bugs me here is that you can find an average rotational velocity from t=10 s to
    t =18 s, but acceleration should be calculated from instantaneous velocities. I am guess you are expected to figure that the average rotational velocity occurs at the center of each interval...
     
  11. May 25, 2012 #10
    Conceptualize.
    If a diagram is not provided, you should almost always make a quick drawing of the situation.
    Shouldn't rely on equations for plug and chug.

    For a constant acceleration kinematic, you should be able to draw and interpret a velocity-time diagram.
     
  12. May 26, 2012 #11
    Can I do by this
    Q = w0t + 1/2 at^2
    200 = 10W0 + 50a ====== first equation


    Q = w0t + 1/2 at^2
    100 =8w0 + 32a ======== second equation

    200 = 10W0 + 50a
    100 =8w0 + 32a
    -------------------------
    100 = 2W0 + 18a


    is this correct until now ?
     
  13. May 26, 2012 #12
    Segment #2
    Constant velocity->8 secs -> 100reve.
    Constant velocity?
    Equation?

    Segment #1
    Constant acceleration ->10 secs -> 200 rev.
    Equation?
     
    Last edited: May 26, 2012
  14. May 26, 2012 #13
    see the Q2 ... It want initial velocity during the acceleration was constant and after change to the a = 0 as the velocity is constant

    now I will ask is the initial velocity depend for constant acceleration or from it was constant to changed .

    if the initial velocity depend only to first rev that mean
    w = w0 + at
    w = 20 rad/s
    but now we have 2 variable a and w0

    How I can get one of them ????
     
  15. May 26, 2012 #14
    What is the value of constant velocity?

    This value is the v of constant acceleration.

    s=ut+1/2at2
    a=(v-u)/t

    s=ut+1/2t2(v-u)/t

    2s=2ut + vt-ut

    u=(2s-vt)/t
     
    Last edited: May 26, 2012
  16. May 26, 2012 #15
    constant velocity is 12.5 rad/s
     
  17. May 26, 2012 #16
    OMG !! from where you got this equations

    We learn only 3 equation
    w = w0 + at
    w2 = w0^2 + 2sQ
    Q = W0t + 1/2 at^2

    I am confusing ??
     
  18. May 26, 2012 #17
    These 2 are standard.
    s=ut+1/2at2 ...(1)
    a=(v-u)/t .....(2)

    Now you substitute "a" in equation (1)
    And you get
    s=ut+1/2t2(v-u)/t

    Just replace the translational variables with rotational variables.
    u ->ω0
    v ->ω1
    a ->α
    s->θ
     
    Last edited: May 26, 2012
  19. May 26, 2012 #18
    why substitute ... why we don't take value for each variables ..

    w = w0 + at
    a = (12.5 - 20 ) 8
    = -0.935


    is correct ..?
     
  20. May 26, 2012 #19
    ω1=12.5 -Correct.
    Where do get the value ω0=20?

    Now we are talking on acceleration NOT equal to zero.
     
  21. May 26, 2012 #20
    12.5 - 20 / 18

    a =-0.4

    is correct ?

    give me the way to find a ? ??????????????????????????????????????????????????????
     
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