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Constant acceleration kinematics

  1. Feb 17, 2015 #1
    In cases like projectile motion, although the acceleration vector is constant (ignoring air resistance), a single coordinate axis fails to describe the motion of the particle, since it occurs in a plane; therefore we need a minimum of two coordinate axes to fully describe the trajectory of that particle. I was wondering, is there a case of constant acceleration where two axes would be insufficient to describe the trajectory (forcing us to use a three dimensional coordinate system)?
     
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  3. Feb 17, 2015 #2

    Suraj M

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    Maybe you could consider the trajectory situation in the presence of wind ➡ in the direction parallel to the ground and perpendicular to the original path of the trajectory!
     
  4. Feb 17, 2015 #3
    The particle would deviate from its original trajectory along a curve and the motion would no longer occur in a single plane?
     
  5. Feb 17, 2015 #4

    Suraj M

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    Exactly, it would look like a normal trajectory when viewed from the zy plane and xy plane. hence you need a 3D representation!
     
  6. Feb 17, 2015 #5

    jtbell

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    But would the acceleration still be constant (in both magnitude and direction) in such a situation?
     
  7. Feb 17, 2015 #6

    PeroK

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    If the acceleration vector is constant (both magnitude and direction), then I'd take that to be my -y axis!

    How would you choose your x-axis?
     
  8. Feb 18, 2015 #7
    If at t=0 the particle experiences two constant forces then the acceleration vector would be constant and along the resultant of the two forces, right?
     
  9. Feb 18, 2015 #8

    jtbell

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    The force from air drag depends on the relative velocity of the object and the air.
     
  10. Feb 18, 2015 #9
    I suppose this wasn't a very good example then. Back to my original question: is there a situation where a plane would cease to fully describe a trajectory where the acceleration is constant?
     
  11. Feb 18, 2015 #10

    A.T.

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    I don't think so. The plane containing the initial velocity and the constant acceleration vectors can never be left by the object.
     
  12. Feb 18, 2015 #11

    jtbell

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    Suppose an object has initial velocity ##\vec v_0## and constant acceleration ##\vec a##. During a time interval ##\Delta t_1## it changes velocity by ##\Delta \vec v_1 = \vec a \Delta t_1## and has a new velocity ##\vec v_1 = \vec v_0 + \Delta \vec v_1##. The three vectors in this equation lie in a plane.

    The object continues during a second time interval ##\Delta t_2##, changes its velocity by ##\Delta \vec v_2 = \vec a \Delta t_2##, and has a new velocity ##\vec v_2 = \vec v_1 + \Delta \vec v_2##.

    Given that ##\vec a## is constant (in both magnitude and direction), can ##\Delta \vec v_2## and ##\vec v_2## lie outside the plane defined in the first paragraph?
     
  13. Feb 18, 2015 #12

    PeroK

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    Another way to look at it, is to take one axis parallel to the acceleration and the other parallel to the component of the velocity perpendicular to the acceleration. That reduces things to 2D motion.

    Even with a central force, you can choose the coordinates so that motion is 2D: choose the y-axis along the initial radial vector; and choose the x-axis parallel to the component of the initial velocity perpendicular to that.
     
  14. Feb 18, 2015 #13
    ##\Delta \vec{v}_2## and ##\vec{a}## are collinear, so ##\Delta \vec{v}_2## and ##\Delta \vec{v}_1## lie in the same plane. Adding them together yields a vector which has to lie in the same plane they are in. Therefore, all vectors lie in the same plane. Right?
     
  15. Feb 18, 2015 #14

    Suraj M

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    I didn't say air drag, i said wind in the horizontal direction perpendicular to the path of a normal trajectory! wouldn't that be in 3D?(OP's original question)?
     
  16. Feb 18, 2015 #15
    The point is that with any constant wind flow, regardless of direction, a resultant acceleration vector can be found out (resultant of the acceleration due to wind flow and due to gravity), and this would again result in motion that can be described in a single plane. There is no global coordinate system here, so you can just arbitrarily redefine your x and y axes to describe the motion in the xy plane.
     
  17. Feb 18, 2015 #16

    Suraj M

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    I'm sorry Pwiz but i'm not able to imagine the 3D situation. This is my interpretation, its probably wrong, but whats wrong in it?Could you point it out?
    WIN_20150218_224244.JPG
     
  18. Feb 18, 2015 #17
    That is because the situation cannot be said to be in 3D if you redefine your axes. Try imagining that the object makes a parabolic path between the x, y and z axes somewhere. Now what if you "rotated" your view so that the parabolic path faces you? Would the motion still be 3 dimensional?
     
  19. Feb 18, 2015 #18

    jbriggs444

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    An important point being that the wind flow must be constant relative to the projectile's current velocity so that the resultant of gravity plus wind resistance will be constant, correct? The OP required constant acceleration. That entails that the wind velocity in the ground frame would need to be increasing at a steady pace both horizontally and downward so as to maintain a constant relative speed and, accordingly, a constant force on the projectile.

    If the wind speed were held constant in the ground frame, the 3D trajectory would end by asymptotically approaching a line in a fixed direction depending only on gravity and wind and would start on a line in the direction of its starting velocity. Those two lines need not be parallel. They could be skew, thus not fitting into a single plane.
     
  20. Feb 18, 2015 #19
    By a constant wind flow, I meant a wind which exerts a constant force on the object(doesn't change direction as well), since only then can the resultant acceleration vector have a constant direction and magnitude resulting in planar motion, as you have pointed out.
     
    Last edited: Feb 18, 2015
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