physics help

The magnitude of the velocity of a particle
which starts from rest 2 ft below the origin when
t = 0 and moves along a vertical axis is directly
proportional to the time after starting. The
displacement of the particle during the time
interval from t = 1 sec to t = 2 sec is 3 ft upward.
When t = 3 sec, determine (a) the location of the
particle; (b) the acceleration of the particle.

asked by stan
  1. Hey Stan, is that a direct copy/paste? Ie that's all your were given word for word? Hoi.. I think my Physics is rusty.

    posted by EssKay
  2. yeah that is word for word

    posted by stan
  3. Hey again Stan, hmm, okay, I'll keep playing with it, I still say we're missing info though... but re-post in the meantime, so other tutors don't think I've answered.

    posted by EssKay
  4. Acceleration is constant. Initial velocity is zero. You know that from the first sentence.

    Vertical location is
    Y = -2 + (a/2)t^2
    Y(t=2) - Y(t=1) = 3 ft.
    (a/2) (2^2 - 1^2)) = (3/2)a = 3
    a = 2 ft/s^2
    Y(t=3) = -2 + (2/2)*3^2 = 7 feet

    posted by drwls

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