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.

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

yeah that is word for word
posted by stan

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

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 feetposted by drwls
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