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.