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X versus t graph, simple kinematics?

  1. Sep 29, 2004 #1
    can someone help me graph this?

    http://www.physics.drexel.edu/courses/Physics-152/probgraph.gif

    I can understand that, since the velocity is changing, the x versus t graph should not be a straight line, that it is curved, but that's about all I can figure out.

    I don't have a textbook, because the bookstore is backordered on all of them, plus this was not covered in the first lecture, so all I can figure out is that it's some curve...

    Please help me... :confused:
     
  2. jcsd
  3. Sep 29, 2004 #2
    from 0 to 50 the object speeds up. from 50 to 90 it slows down. from 90 to 105 it is at rest. 105> it speeds up.... ???
     
  4. Sep 29, 2004 #3
    yeahh, some of us were confused too, and did what the instructions on the pic show, but we're supposed to graph an x versus t graph of that picture showing the velocity, so it looks somewhat like an exponential growth graph...
     
  5. Sep 29, 2004 #4
    that's all I can get, just that it curves up from 0 but then I know there are other dips and peaks, and a level off since the velocity goes to zero, but I don't know how to draw it
     
  6. Sep 29, 2004 #5
    what does it ask you to find? accelaration or distance?

    accelaration is speed/ time and distance is speed * time..
     
  7. Sep 29, 2004 #6
    the question was this exactly...

    "7 Chapter 2, Question 22,and plot an x vs t graph of the motion."

    so I mean, the directions aren't that helpful, plot an x vs t graph of the motion
     
  8. Sep 29, 2004 #7
    the book I have does have a drawing of a x versus t graph, but it derives it from already having acceleration... and used the equation

    x=Initial Velocity x Time + 1/2 x Acceleration x Time^2
     
  9. Sep 29, 2004 #8
    ok so it is distance vs time

    distance = speed * time

    from that graph speed = y-axis and time= x-axis.
     
  10. Sep 29, 2004 #9
    wait, now you lost me... how does that help me graph something in

    x intervals of Time t (s) and y intervals of Position x (m)
     
  11. Sep 29, 2004 #10
    yes that is the equation for distance.....

    from the graph above acceleration= [V(final)-V(initial)]/[t(final)-t(initial)]
    in other words a= (V2-V1)/(t2-t1)
    initial velocity= 15m/s

    just graph velocity*time from the given graph above
     
  12. Sep 29, 2004 #11
    so it's (10-15)/(120-0) to get acceleration of -1/24

    and I just graph

    15x + (1/2)(-1/24)x^2? all I got was a straight line
     
  13. Sep 29, 2004 #12
    couldn't the final velocity be any point along the line in the above graph? see that's whats confusing me the most, that there aren't any concrete numbers that I can see to use

    I mean, for the first part of the graph, I can see yes, the equation should be

    y=1/2x+15

    because the slope is 1/2 and picking a point (10,20) plugging that in, got me the y intercept... but that just happened to work out
     
  14. Sep 29, 2004 #13
    but that y intercept equation, I don't see how it helps, it just gives me an equation for a line I already saw, but not even the whole line...
     
  15. Sep 29, 2004 #14
    d= v0xt-0.5 x a x t is good for uniform acceleration

    so use d= (v2-v1)* 0.5 (t2-t1)

    so
    v2=20
    v1=12
    t2=10
    t1=0
    d1=0.5*(20-12)*(10-0)=40

    so you graph (10,40)

    next

    v2=25
    v1=20
    t2=20
    t1=10
    d2=0.5*(25-20)*(20-10)=25
    total distance= d1+d2=40+25=65

    graph (20,65)

    and so on....

    I hope this is right otherwise I'll go insane.
     

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    Last edited: Sep 29, 2004
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