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.
4 answers
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.
yeah that is word for word
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.
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
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
1 answer