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Homework Help: Triangular numbers and v-t graph

  1. May 30, 2005 #1
    http://student.shcc.edu.hk/~s021107/phy.GIF [Broken]

    Displacement=area under v-t graph = 150m

    using another concept,
    the acceleration is 0.75m^s-2, so the displacement at
    1s=0.75m
    2s=0.75*2m
    .
    .
    .
    20s=0.75*20m

    total displacement =sum of displacement of every second
    =0.75*(1+2+3+...+18+19+20)
    =157.5m
    Can you tell me wt's wrong here?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 30, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your link isn't working, but I assume you are describing uniformly accelerated motion starting from rest.

    Since [itex]v = at[/itex]:

    At t = 0s, the speed is 0 m/s
    At t = 1s, the speed is 1*.75 = 0.75 m/s
    At t = 2s, the speed is 2*.75 = 1.5 m/s, .... etc.

    To find the displacement during each second, use [itex]\Delta d = v_{ave} \Delta t[/itex], where the average speed equals [itex]v_{ave} = (v_i + v_f)/2[/itex]. Add these displacements and you'll find that they total to 150m.
     
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