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Position vs Time. Need help

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Position vs Time graph is given. It is a perfect parabola. So it has a quadratic equation
    Y: Ax^2+Bx+C that fits it perfectly.

    (a) Determine acceleration, initial velocity, initial position
    (b) Sketch a motion map
    (c) Sketch a v vs t graph

    My attempt:
    initial position: 50 m off the ground (along y axis) from the graph t=0 sec here
    initial velocity: slope of the line b/w first and second point?
    53-5-50/1 (50 initial position; 53.5 position at t=1 sec and from here on the object goes into free fall)
    acceleration: change in velocities/ time taken I'm confused about this because velocity will be 0m/s at 1 sec because the object is about to go into free fall. How do I calculate this?

    (b) & (c) I don't have a clue what motion map is and v vs t can be drawn using position vs time

    Thanks!
     
  2. jcsd
  3. Feb 16, 2014 #2
    You have the initial position correct, but for initial velocity, it needs to be the slope of the line at t = 0, so what tool allows to find instantaneous slopes? And for acceleration, you can either you this same tool twice on the position function, or, since it's in free-fall, what do you know about the object's acceleration?

    A motion map is (I'm assuming) the same thing as a path equation. This means you want to graph the motion on the x-y plane instead of the x-t plane. That is, graph an actual trace of the object's motion as an observer would see it.
     
  4. Feb 16, 2014 #3
    At t=0, the object is at rest. The person hasn't thrown it away yet. I'll give you some of the values from my graph.
    time (s) position (m)
    0 50
    1 53.2
    2 52
    3 45.5
    4 34
    5 17.5
    6 -4.0

    So, if I'm understanding it right. Since it is a 2D motion, it was thrown with some force. I need to find that initial velocity. The mass of the object isn't given, neither is the height of the cliff. So we can't really use equations. The object is the the peak of parabola at t=1. From here on, it's a free fall. The acceleration will be -9.8 m/s^2. What about the acceleration in the duration t=0 and t=1?
    Slope of the line t=0
    53.5-50/1 [delta d/delta t] In other words, displacement/time
    3.5 m/s <--- is that my initial velocity?
    will the acceleration up to this point (t=1) be 0 since the velocity remains the same? and from this point onwards the velocity is constantly changing as it is acted upon by g?
     
  5. Feb 16, 2014 #4
    Oh, I didn't realize you were given a graph without the equation, so really every thing is just going to be an approximaton. For the initial velocity, it won't be zero at t = 0, since that's assumed to be the start of the motion, and it has to have some velocity to start. What you have is one way to get a value for it, but even better would be to draw a tangent line at 0 and see what the slope of that line is.

    I don't follow your reasoning with the acceleration from t = 0 to 1. Shouldn't the velocity have to change from the initial velocity, in which case it would be nonzero?
     
  6. Feb 16, 2014 #5
    Perhaps that's why we're only asked to sketch the graphs because it is only an approximation. Also, which would the ideal equation to describe the motion?

    I am thinking
    S= vt + 1/2 at^2
     
  7. Feb 16, 2014 #6
    Yes, that would work. Keep in mind though that s really means the change in position, so you'd want to add in the initial position to the right side. Also, make sure v is the initial velocity, and that should be good.
     
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