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Calculating distance per 1-sec interval for plot graph d vs t

  1. Sep 28, 2011 #1
    2 travelining in a straight line initial velocity is 14m/s, at a constant acceleration of 2.0m/s^2 to a velocity of 24 m/s

    a) how much time to reach velocity of 24m/s

    t= v-v0/a or 24-14/2= 5 seconds

    b) distance covered by car in this process?

    d=v0(t) + 1/2(at^2) or (14)(5)+1/2(2)(5^2)= 70m+25m= 95m

    c) compute values of the distance traveled at 1-sec intercals and carefully draw a graph of distance plotted against time for this motion?
    c is the only one throwing me off

    the problem doesnt say it is constant but this is the approach I took to form a graph plot

    so then distance per second would be 19 m?
    by taking distance/time 95m/5s
    so at each second interval it would be like this t graph?

    t-0 d-0
    t-1 d-19
    t-2 d- 38

    so on and so forth?
    I mean it doesnt say its constant acceleration so idk if what I did was right for c
     
  2. jcsd
  3. Sep 28, 2011 #2

    berkeman

    User Avatar

    Staff: Mentor

    On a), I don't think you used the equation correctly. Start with the equation for velocity and carefully apply your numbers...

    [tex]v(t) = v_0 + at[/tex]
     
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