Question

What is the acceleration of a ball rolling 12 m/s to a stop in 3 seconds?
answers:

-0.25 m/s2

-4m/s2

4m/s2

0.25 m/s2

Answer 1

To find the acceleration of the ball, we can use the formula for acceleration:

\[
a = \frac{\Delta v}{\Delta t}
\]

where:
- \(\Delta v\) is the change in velocity,
- \(\Delta t\) is the change in time.

In this case:
- The initial velocity \(v_i = 12 \, \text{m/s}\) (the ball's velocity before stopping),
- The final velocity \(v_f = 0 \, \text{m/s}\) (the ball's velocity after stopping),
- The time taken to stop \(\Delta t = 3 \, \text{s}\).

Now, calculate \(\Delta v\):

\[
\Delta v = v_f - v_i = 0 \, \text{m/s} - 12 \, \text{m/s} = -12 \, \text{m/s}
\]

Now plug this into the acceleration formula:

\[
a = \frac{-12 \, \text{m/s}}{3 \, \text{s}} = -4 \, \text{m/s}^2
\]

So the acceleration of the ball is \(-4 \, \text{m/s}^2\).

The correct answer is **-4 m/s²**.