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Velocity to Distance (graph)

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A velocity of a runner is described in the graph below. The runner started from X=0 on the X axis. In what distance from X=0 (beggining) will be the runner, after 11.5 sec.

    2hqcewm.jpg


    2. Relevant equations



    3. The attempt at a solution

    Please tell me why it isnt 42m.:confused:
     
  2. jcsd
  3. Nov 8, 2012 #2
    V=m*s, so that means the area of the graph gives us the distance from X=0. then;

    32(the 2*1 areas)+ 4 (the first triangle) + 2 (the second triangle) (from X=0 to 10. sec)

    now calculate the +1,5 sec area which is 1,5*4=6

    then 38+6=44, right?
     
    Last edited: Nov 8, 2012
  4. Nov 8, 2012 #3
    Thanks goktr001, but i have just solved it, its actually 44.
    I summed the area under the graph, but only till the 11sec, i didnt notice that 11.5 is actually 3 quarders of triangle.

    Problem solved anyway. thanks.
     
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