Question

what is the acceleration of a ball rolling 12 m/s to a stop in 3 seconds? (1 point)
4 m/s^2
0.25 m/s^2

• 4 m/s^2
• 0.25 m/s^2

Explanation:

Step1: Identify initial - final velocities and time

Initial velocity $u = 12$ m/s, final velocity $v = 0$ m/s, time $t=3$ s.

Step2: Apply acceleration formula

The acceleration formula is $a=\frac{v - u}{t}$. Substitute the values: $a=\frac{0 - 12}{3}$.

Step3: Calculate the acceleration

$a=\frac{- 12}{3}=-4$ m/s².

Answer:

C. - 4 m/s²