A ball rolling down a hill was displaced at 19.6m while uniformly accelerating from rest. If the final velocity was 5.00 m/s what was the rate of acceptation?
3.5 m/s root 2
To find the rate of acceleration, we can use the following equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity = 5.00 m/s
vi = initial velocity = 0 m/s (since the ball was at rest)
d = displacement = 19.6 m
Substituting the given values into the equation:
(5.00 m/s)^2 = (0 m/s)^2 + 2a(19.6 m)
25 m^2/s^2 = 39.2a
To solve for 'a', divide both sides by 39.2:
a = 25 m^2/s^2 / 39.2
a ≈ 0.6378 m/s^2
Therefore, the rate of acceleration is approximately 0.6378 m/s^2.
To find the rate of acceleration, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity = 5.00 m/s
vi = initial velocity = 0 m/s (since the ball starts from rest)
d = displacement = 19.6 m
a = acceleration (unknown)
Rearranging the equation to solve for acceleration:
a = (vf^2 - vi^2) / (2d)
Substituting the given values:
a = (5.00^2 - 0^2) / (2 * 19.6)
Now, let's calculate the value:
a = (25 - 0) / 39.2
a = 25 / 39.2
a ≈ 0.6378 m/s^2
Therefore, the rate of acceleration for the ball rolling down the hill is approximately 0.6378 m/s^2.
a = (V^2-Vo^2)/2d
a = (5^2-0)/39.2 = 0.6378 m/s^2.