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Help with 2d motion

  1. Sep 26, 2004 #1
    2. [PSE6 4.P.005.] At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.80 s, the particle's velocity is v = (9.10 i + 8.50 j) m/s.
    (a) Find the acceleration of the particle at any time t. (Use t, i, and j as necessary.) Got this one


    (b) Find its coordinates at any time t.
    x = ( ) m
    y = ( ) m

    Don't get these two.
     
  2. jcsd
  3. Sep 26, 2004 #2

    Pyrrhus

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    The problem states constant acceleration, so use the kinematics equations for uniform acceleration, with both the speeds (x-axis and y-axis) and the acceleration(x-axis and y-axis) you found to find the coordinate on the x-axis, and then on the y-axis.
     
  4. Sep 26, 2004 #3
    One other question:

    To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 330 m/s at 46.0° above the horizontal. It explodes on the mountainside 36.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?

    I got the x coordinates by doing 330 cos 46 * 36, but when I try yf = 330sin46 - 1/2(-9.8)(36)^2, the answer is wrong. How do I find the y coordinates?
     
  5. Sep 26, 2004 #4

    Pyrrhus

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    You forgot time on your initial speed
    :smile:

    [tex] Y - Y_{o} = V_{o}t + \frac{1}{2}at^2 [/tex]
     
  6. Sep 26, 2004 #5
    Well, I tried yf = 330sin46(36) - 1/2(-9.8)(36)^2 and it turned out to be about 14896 feet, which was wrong lol, any ideas?
     
  7. Sep 26, 2004 #6

    Pyrrhus

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    You already has a negative sign for acceleration, it seems to me you're plugging it twice.
     
  8. Sep 26, 2004 #7
    oo okay, I just tried yf = 330sin46 + 1/2(-9.8)(36)^2, and got 2195 feet, that sound right?
     
  9. Sep 26, 2004 #8

    Pyrrhus

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    feet? you mean meters, It's the answer yes.
     
  10. Sep 26, 2004 #9
    lol thanks, but yeah, one other one that was giving me problems: PSE6 4.P.014.] An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 10.0 m if her initial speed is 2.40 m/s. What is the free-fall acceleration on the planet?

    I plugged stuff into the kinematics formula, but not getting it
     
  11. Sep 26, 2004 #10
    O never mind
     
  12. Sep 26, 2004 #11
    6. [PSE6 4.P.029.] A tire 0.300 m in radius rotates at a constant rate of 190 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2r.)
    m/s
    What is the acceleration of the stone?

    lol yeah, need help with this one
     
  13. Sep 26, 2004 #12

    Pyrrhus

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    Think about the hint, and you will get it.
     
  14. Sep 26, 2004 #13
    Well I tried 2pi(.300), but I'm wondering if you have to convert the rev/minute
     
  15. Sep 26, 2004 #14

    Pyrrhus

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    Let me phrase it this way

    1 revolution is 2pir meters.

    and you got 190 rev/min, so how many meters will be 190 revs?
     
  16. Sep 26, 2004 #15
    2pi(190) i get 1193.8 meters
     
  17. Sep 26, 2004 #16

    Pyrrhus

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    maybe you didn't see the r for radius

    1 rev is 2piR meters

    190 rev is ????
     
  18. Sep 26, 2004 #17
    oo okay, I solved 190=2pir and got 30.239 m
     
  19. Sep 26, 2004 #18
    I divided that by 60 to get .504 m/s, where do I go from here?
     
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