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.
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