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.

1 answer

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 \( 1.67 , \text{m/s}^2 \).