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For no apparent reason, a poodle is running counter-clockwise at a constant speed of 3.40 {\rm m/s} in a circle with radius 2.2 {\rm m}. Let \vec v_1 be the velocity vector at time t_1, and let \vec v_2 be the velocity vector at time t_2. Consider \Delta \vec v = \vec v_2- \vec v_1 and \Delta t = t_2 - t_1. Recall that \vec a_{\rm av} = \Delta \vec v/ \Delta t.

  • physics -

    Similar problem done for Jimmy, note different numbers for speed and radius:

    geometry if a = 0 at t = 0
    angle from x axis (call it a) = v t/r

    Vx = -v sin a
    Vy = v cos a

    at start a = 0 so
    V1x = 0
    V1y = v
    at time t
    V2x = -v sin(vt/r)
    V2y = v cos(vt/r)

    call change d

    d Vx = -v sin(vt/r)
    d Vy = v[cos(vt/r)-1]

    ax = -(v/t) sin(vt/r)
    ay = (v/t)[cos(vt/r)-1]

    so for our numbers
    v/t = 5.4/.7 = 7.714
    vt/r = 1.643 about 94 degrees, quadrant2
    ax = -7.714 sin(1.643) = -7.694
    ay= 7.714[cos(1.643)-1] = -8.270
    |A| =SQRT(ax^2+ay^2) = 11.30 m/s^2

    compare to v^2/r
    v^2/r = 5.4^2/2.3 = 12.7 not too far off centripetal acceleration

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