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Help with graphs

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Three particles move along the x-axis, each starting with v= 10m/s at t_0 = 0s. The graph for A is a position-versus-time graph; the graph for B is a velocity-versus-time graph; the graph for C is an acceleration-versus-time graph.

    Find each particle's velocity at t = 7.0s. Work with the geometry of the graphs, not with kinematic equations.
    http://img27.imageshack.us/img27/5497/phys.jpg [Broken]

    2. Relevant equations
    A=1/2(b)(h)
    A=bh
    3. The attempt at a solution

    Particle A:
    A=1/2(5)(50)
    =125m/7s
    =-17.85m/s, negative because the particle is moving leftwards.

    Particle B:

    Since this is a Velocity vs. Time graph, the Velocity axis shows the object to be moving at 20m/s.

    Particle C:

    At t=2s, the car stops(?) and begins to accelerate till t=8s. At t=7s, would the acceleration be 50m/s^2?

    Can anyone check if I'm doing these correctly at all, and if not can you advise me, please?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 9, 2009 #2
    A:

    Not quite. Velocity is the derivative of position - its rate of change. Find the slope of the graph at t=7 and you've got the particle's velocity.

    B:

    Double check that answer. Very small error, but you've got the right idea. You do just need to find the graph's value at t=7, but take another look at what that value is.

    I'm not totally sure how to do C at the moment, I'll keep thinking.
     
  4. Sep 9, 2009 #3
    A. Displacement is area under the curve ie, from the line to the x axis. So you would find the areas of the larger odd shape, and the take away the area of the small triangle then divide by t. OR think about what the slope is. it's [tex]\Delta y / \Delta x[/tex] In this case y is displacement, and x is time. So that means that the slope would be what?

    B. Yes

    C. In an a/t graph, area under the curve would be velocity change.
     
  5. Sep 9, 2009 #4
    Ok, so:
    A) -10m/s
    B) -20m/s

    I'm still a bit confused by C, do you just mean the area just under t=2s to t=7s?
     
  6. Sep 9, 2009 #5
    A and B are correct, I'm still not sure on C.
     
  7. Sep 9, 2009 #6
    Oh, I got it. I found the triangle area from t=2s to t=7s. Then I found the area of the rectangle underneath it. Which ended up being 125-50 = 75m/s.

    Thanks for the help!
     
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