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Velocity time graph find position, acce?

  1. Sep 23, 2010 #1
    The velocity-versus-time graph is shown for a particle moving along the x-axis. Its initial position is xo=4.80 m at to=0.00 s. What is the particle's position at t=1.25 s if the total time the particle moves is 5.00 s and the maximum velocity is 2.00 m/s?


    1)At t=3.75 s, what is the particle's position?

    2)At t=3.75 s, what is the particle's acceleration?

    I tried to find the position by finding the area up until 3.75,but the answer seems not right...
    and acceleration i used v final - v initial / change in time. but im not sure if it's right or not..

    if some1 can help with the steps, that'll be great~thanks
     

    Attached Files:

    Last edited: Sep 23, 2010
  2. jcsd
  3. Sep 23, 2010 #2
    Acceleration is the slope of a velocity-time graph.
    Change in position is the area under the graph from the start time to the end time. - Remember to add this to the initial position

    Until you do calculus, you pretty much need to learn that by heart. (If you have some calculus, the above basically says a = dv/dt and that change in position is the integral of velocity.)

    Slope = change in y/change in x
    You need to calculate this for a straight portion of the graph. If there are kinks in the graph, it means the acceleration is changing. You need to work on a part where it's constant.
     
  4. Sep 23, 2010 #3
    got it thanks.
     
  5. Sep 24, 2010 #4
    how would you find the area for last small triangle 3-3.75s
    1/5(0.75)(what would be height ?
     
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