A ball rolls down a hill with a constant acceleration of 2.0 m/s2. If the ball starts from rest, (a) what is its velocity and the end of 4.0 s? (b) How far did the ball move?
1.23m/s
To solve this problem, we can use the equations of motion. There are three equations that relate displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t):
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
(a) To find the velocity at the end of 4.0 seconds, we can use equation 1. Here, the initial velocity, u, is 0 m/s since the ball starts from rest, the acceleration, a, is 2.0 m/s^2, and the time, t, is 4.0 s. Plugging these values into equation 1, we have:
v = 0 + (2.0)(4.0)
v = 0 + 8.0
v = 8.0 m/s
Therefore, the velocity of the ball at the end of 4.0 seconds is 8.0 m/s.
(b) To find the distance the ball moved, we can use equation 2. Plugging in the values we know, u = 0 m/s, a = 2.0 m/s^2, and t = 4.0 s, we have:
s = (0)(4.0) + (1/2)(2.0)(4.0)^2
s = 0 + (1/2)(2.0)(16.0)
s = (1/2)(2.0)(16.0)
s = (1.0)(16.0)
s = 16.0 m
Therefore, the ball moved a distance of 16.0 meters.