Question
A hoop rolls down a hill. In 6.0 6.0 seconds, it goes from 2.0 m/s 2.0 m / s to 12 m/s 12 m / s . In 1–2 sentences, calculate the acceleration of the hoop. Show or explain your work.
Answers
GPT-4o mini
To calculate the acceleration of the hoop, you can use the formula for acceleration:
\[
a = \frac{\Delta v}{\Delta t}
\]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time interval.
Here, the initial velocity \( v_i = 2.0 \, \text{m/s} \) and the final velocity \( v_f = 12.0 \, \text{m/s} \), so:
\[
\Delta v = v_f - v_i = 12.0 \, \text{m/s} - 2.0 \, \text{m/s} = 10.0 \, \text{m/s}
\]
The time interval \( \Delta t = 6.0 \, \text{s} \). Now, plug this into the acceleration formula:
\[
a = \frac{10.0 \, \text{m/s}}{6.0 \, \text{s}} \approx 1.67 \, \text{m/s}^2
\]
Thus, the acceleration of the hoop is approximately \( 1.67 \, \text{m/s}^2 \).
\[
a = \frac{\Delta v}{\Delta t}
\]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time interval.
Here, the initial velocity \( v_i = 2.0 \, \text{m/s} \) and the final velocity \( v_f = 12.0 \, \text{m/s} \), so:
\[
\Delta v = v_f - v_i = 12.0 \, \text{m/s} - 2.0 \, \text{m/s} = 10.0 \, \text{m/s}
\]
The time interval \( \Delta t = 6.0 \, \text{s} \). Now, plug this into the acceleration formula:
\[
a = \frac{10.0 \, \text{m/s}}{6.0 \, \text{s}} \approx 1.67 \, \text{m/s}^2
\]
Thus, the acceleration of the hoop is approximately \( 1.67 \, \text{m/s}^2 \).