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!

Finding velocity at t=7

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Find each particle's velocity at t=7.0s. Work with the geometry of the graphs, not with kinematic equations.

    2. Relevant equations

    3. The attempt at a solution
    I'm not sure how to arrive at an answer. For the previous question (velocity from graph b) I found the value of -20 m/s by reading the graph and before that (velocity from graph a) I did a simple delta/delta[t] calculation and got -10 m/s. My issue is that I can't find an velocity at t=7.0s for graph C (the acceleration graph.) Please help.
    Thank you.

    Attached Files:

    Last edited: Oct 3, 2012
  2. jcsd
  3. Oct 3, 2012 #2


    User Avatar
    Science Advisor

    Velocity is the integral of acceleration with respect to time. How could you read that off a graph?
  4. Oct 3, 2012 #3
    by integrating the area at that point to the x axis?
    Last edited: Oct 3, 2012
  5. Oct 3, 2012 #4


    User Avatar
    Science Advisor

    Integration is right, presuming you meant to the y axis. I don't think you've got to the right answer though, since there's more than one triangle to worry about in the 0-7s region of the graph. You can check your answer by counting squares for a graph this simple.

    Remember that the integral just gives you the velocity change over the time period, which isn't quite what you're asked for.
  6. Oct 3, 2012 #5
    how would i isolate
  7. Oct 3, 2012 #6
    Find a(t) and do antiderivative to find v(t).
    Last edited: Oct 3, 2012
  8. Oct 3, 2012 #7
    still can't get it.
    i tried calculating the area underneath both of those and got 10 m/s, but that wasn't the velocity of the acceleration graph... im not sure what to do
  9. Oct 3, 2012 #8
    as much as i hate to ask for a straight answer (with a minor explanation) from you guys, i think it would be beneficial because i simply can't get it and i've inputted about 12 answers into this program
  10. Oct 6, 2012 #9


    User Avatar
    Science Advisor

    Complete answers are forbidden by forum rules.

    The area under the graph can be obtained by counting squares between the graph and the x-axis, remembering that the area is negative when the graph is below the axis. Each square is 1s wide and 10ms-2 high. To start you off, the area between 5s and 7s is -20ms-1. You can work out the area between 0s and 5s.

    Do remember that the result is the velocity gain over the period, not the final velocity.
    Last edited: Oct 6, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook