A ball rolling down a hill was displaced 19.6 m while uniformly accelerating from rest. If the final velocity was 6.00m/s what was the rate of acceleration
d = 1/2 a t^2
19.6 = 1/2 * a * 6.00^2
To find the rate of acceleration, we can use the equation of motion:
Δx = (v^2 - u^2) / (2a),
where:
Δx is the displacement,
v is the final velocity,
u is the initial velocity (which is 0 since the ball started from rest), and
a is the acceleration.
In this case, Δx = 19.6 m, v = 6.00 m/s, and u = 0 m/s.
Substituting the given values, we have:
19.6 = (6.00^2 - 0^2) / (2a).
Simplifying the equation gives:
19.6 = 36 / (2a).
To find a, we need to isolate it. We can rearrange the equation as follows:
2a = 36 / 19.6.
Divide both sides by 2:
a = 18 / 19.6.
Evaluating this expression will give us the rate of acceleration:
a ≈ 0.918 m/s^2.
Therefore, the rate of acceleration is approximately 0.918 m/s^2.