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Kinematics of linear motion

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data
    a.jpg
    I am unable to solve part b and please help me check whether is there anything wrong with my graph..:redface:


    3. The attempt at a solution
    graph.jpg

    Segment A
    Vi=0
    V=3
    S=60
    a=9.81

    v=vi+at
    3=0+9.8(t)
    t=0.305s

    Segment B
    Vi=3
    V=0
    a=-2

    v=vi+at
    0=3-2t
    t=1

    t=1+0.305=wrong answer :cry:
     
  2. jcsd
  3. May 16, 2012 #2

    Astronuc

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    Staff Emeritus
    Science Advisor

  4. May 16, 2012 #3
    I think you should draw the diagram roughly to scale.
    The free fall acceleration should be 9.8m/s2 Thus in 1 sec. the velocity should be changing from 0 to 9.8m/s.

    As you see from the graph the area first triangle is equal to 60m.
    From simple formula of triangle you find the height(the velocity)

    From your diagram the time= 40 sec(60=1/2vt). which wrong. After 40 sec. the velocity will be more than 3m/s.
     
  5. May 16, 2012 #4
    azizlwl: So the tip of my triangle should have abit of curve?

    Astronuc: Which of my velocity is incorrect? The 2 website that you listed is for me to check my answer?
     
  6. May 16, 2012 #5
    First try to find all the variables in segment A.
    Acceleration=?
    Displacement=?
    Initial velocity=?
    final velocity =?
    Time taken=?

    What equation should you use to find the final velocity if you have the values of acceleration and displacement?
     
  7. May 19, 2012 #6
    Segment A
    Vi=0
    V=3
    S=60
    a=9.81

    v=vi+at
    3=0+9.8(t)
    t=0.305s

    What equation should you use to find the final velocity if you have the values of acceleration and displacement?

    I would use v^2 = vi^2 +2as
     
  8. May 20, 2012 #7

    Doc Al

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    Staff: Mentor

    I assume segment A is the free fall segment. The 3 m/s is the final speed at the end of segment B--it has nothing (directly) to do with segment A.

    That's the equation you need to figure out the final speed at the end of segment A.
     
  9. May 20, 2012 #8
    Hey Doc Al, thanks for answering :)

    "The parachute opens he experiences a deceleration of 2m/s^2 until he reaches the ground with a velocity of 3m/s"

    Shouldn't it be 3m/s at the start of segment B and Final V = 0 since all things that reach the ground should have 0 speed because it has landed:redface:

    So if you are saying "The 3 m/s is the final speed at the end of segment B" that means my velocity time graph is wrong already.:cry:
     
  10. May 20, 2012 #9

    Doc Al

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    Staff: Mentor

    No. When they say that it reaches the ground with some speed, they mean speed just before it hits. (Once it hits the ground other forces come into play, but we don't care about that motion.)
    That's exactly what I'm saying.

    The first step is to correctly find the speed at the end of segment A (which will also be the speed at the beginning of segment B).
     
  11. May 20, 2012 #10
    Sorry i want to ask a stupid question. How do you see if the question has segments in it?
     
    Last edited: May 20, 2012
  12. May 20, 2012 #11
    gg.jpg

    Doc Al, can you help me take a look at my velocity time graph.. I think it's wrong because it looks abit weird to me :redface: The circle part i don't know where to end it since the question says reaches the ground with a 3/ms So it cannot be 0m/s. And the initial velocity is 0 so my point has to stopped there. Is it right?


    PS: Btw i managed to solved this question thanks to you...:eek:
     
  13. May 20, 2012 #12

    Doc Al

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    Staff: Mentor

    One way is to ask yourself if you can describe the motion with a single acceleration (and thus a single equation). Here you are told that the acceleration changes, so it makes sense to break the motion into constant acceleration segments.
     
  14. May 20, 2012 #13

    Doc Al

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    Staff: Mentor

    Looks fine to me except that you can make it a bit closer to scale. The final speed is 3 m/s, but compare its height above the zero mark with the 34.31 m/s height. Other than that, it's good.
    Yay! Good work.
     
  15. May 20, 2012 #14
    Thanks, thanks alot for taking your time to help me.:approve:
    Thanks to azizlwl & Astronuc too!:tongue:
     
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