1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trajectory Graphs

  1. Sep 21, 2010 #1
    So it is my understanding from my book that the information that we can get from a trajectory graph is the direction of velocity, but not the magnitude of velocity. If the tangent line to a point on a trajectory graph just gives the direction of velocity, why can the trajectory graph still be used to find instantaneous acceleration? Or is the acceleration we find with the trajectory graph just the rate of change of the direction of velocity instead of the rate of change in the magnitude of velocity? My main problem is just understanding how ax, ay, a, v, vx, and vy are related to a trajectory graph. I guess I just can't wrap my head around it because I think that we should only be able to find these values from a position vs. time graph.
     
  2. jcsd
  3. Sep 21, 2010 #2
    Is it just that because you must use the acceleration and velocity in the x direction to find the x position and the acceleration and velocity in the y direction to find the y position that you should be able to find these values from the trajectory graph even though the slope of the trajectory graph only tells the direction of velocity?
     
  4. Sep 21, 2010 #3

    Pythagorean

    User Avatar
    Gold Member

    Well, it depends on what plot you're looking at. If you're looking at the phase space plot, then you only have the x,y positions to see values for.

    You would also plot the nullclines of the system on such a phase plot. Those nullclines (which are lines in a two-dimensional system) tell you exactly where the rate change (dx/dt and dy/dt) are zero. This would be velocity if your system pertains to motion.

    So if your trajectory is on either side of the nullcline, this tells you whether it is a positive rate change or a negative rate change (i.e. a positive or negative velocity would indicate direction). And the farther away from the nullcline, the greater the value of that change rate.

    But because the nullclines are somehow parameterized on the phase plot (i.e. they're not linear with the variables x and y), you can imagine that the real valued change rates would also have to be parameterized on the line. There's no easy way to do that. You would have to plot all the "non-nullcline" lines. And each of the fixed points are going to have different strengths, which you can only analyze by computing the eigenvalues of the jacobian. So this is not so trivial.
     
  5. Sep 21, 2010 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If you want to know the velocity or acceleration, you would need information about time. A graph that has only x and y (and perhaps z) values does not have time information.
     
  6. Sep 21, 2010 #5
    so the graph can't show time, but isn't time a parameter in the position equations for the x and y positions? so if i was given at least 2 or 3 of the variables in the position equation for the horizontal and vertical direction i could find the unknown variable using the position equation that described the trajectory graph right?
     
  7. Sep 22, 2010 #6

    Pythagorean

    User Avatar
    Gold Member

    That would depend on the equation. It may be that a numerical approximation is the only way to solve it.
     
  8. Sep 22, 2010 #7

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    In your first post, you said nothing about having any equations, or any time values; only a graph of x & y. If that graph (with no times indicated, and no equations in terms of time) is all you have, you could not determine the velocity.

    If you are given time values for points along the graph, or are given the equations explicitly in terms of the time, then you could get the velocity.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook