Respuesta :
Answer:
Approximately [tex](-0.33)\; {\rm m\cdot s^{-2}}[/tex] ([tex](-1/3)\; {\rm m\cdot s^{-2}}[/tex]) on average.
Explanation:
Acceleration is the rate of change in velocity.
For the ball in this question:
- Initial velocity: [tex]u = 12\; {\rm m\cdot s^{-1}}[/tex].
- Final velocity: [tex]v = 0\; {\rm m\cdot s^{-1}}[/tex] since the ball has stopped.
- Time required: [tex]\Delta t = 36\; {\rm s}[/tex].
Subtract the initial velocity [tex]u[/tex] from the final velocity [tex]v[/tex] to find the change in velocity:
[tex]\begin{aligned}(\text{change in velocity}) &= (\text{final velocity}) - (\text{initial velocity}) \end{aligned}[/tex].
[tex]\begin{aligned}\Delta v &= v - u \\ &= 0\; {\rm m\cdot s^{-1}} - 12\; {\rm m\cdot s^{-1}} \\ &= (-12)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
(Note that the change in velocity is negative because the final velocity [tex]v = 0\; {\rm m \cdot s^{-1}}[/tex] is more negative than the initial velocity [tex]u = 12\; {\rm m\cdot s^{-1}}[/tex].)
To find the average acceleration [tex]a[/tex] (average rate of change in velocity,) divide the change in velocity [tex]\Delta v[/tex] by the time [tex]\Delta t[/tex] required to achieve such change:
[tex]\begin{aligned}(\text{average acceleration}) &= \frac{(\text{velocity change})}{(\text{time required for change})}\end{aligned}[/tex].
[tex]\begin{aligned} a &= \frac{\Delta v}{\Delta t} \\ &= \frac{(-12)\; {\rm m\cdot s^{-1}}}{36\; {\rm m\cdot s^{-1}}} \\ &\approx (-0.33) \; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
(Average acceleration is negative since velocity is becoming less positive.)