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?
See previous post: Thu,10-10-13,7:46 PM.
6.49 m/s
To find the rate of acceleration, we can use the kinematic equation:
\[v^2 = u^2 + 2as\]
Where:
v = final velocity = 5.00 m/s
u = initial velocity = 0 m/s (as the ball was at rest)
s = displacement = 19.6 m
a = acceleration (what we need to find)
Let's rearrange the equation to solve for acceleration (a):
\[a = \frac{v^2 - u^2}{2s}\]
Substituting the given values:
\[a = \frac{5.00^2 - 0^2}{2 \times 19.6}\]
Simplifying:
\[a = \frac{25.00}{39.2}\]
Calculating:
\[a \approx 0.637 \, m/s^2\]
Therefore, the rate of acceleration for the ball rolling down the hill is approximately 0.637 m/s².