Question 23
1 pts
You travel in circle of radius $5 \mathrm{~m}$. Initially, you are moving at a speed of $5 \mathrm{~m} / \mathrm{s}$ in what we will take to be the positive direction. 2 seconds later, you are at the diametrically opposite point on the circle, again moving at $5 \mathrm{~m} / \mathrm{s}$. What is your average acceleration over this time interval? $\mathrm{m} / \mathrm{s}^{2}$

Step 1 ：Given that the initial velocity is \(v = 5 \, m/s\) and the time interval is \(t = 2 \, s\).

Step 2 ：The average acceleration is defined as the change in velocity divided by the change in time. In this case, the magnitude of the velocity remains the same, but the direction changes by 180 degrees. Since acceleration is a vector quantity, this change in direction implies a non-zero acceleration.

Step 3 ：The change in velocity is twice the initial velocity (since it changes direction completely), and the time interval is 2 seconds.

Step 4 ：Therefore, the average acceleration can be calculated as follows: \[a_{avg} = \frac{\Delta v}{\Delta t} = \frac{2v}{t}\]

Step 5 ：Substituting the given values into the formula, we get \[a_{avg} = \frac{2*5}{2} = 5.0 \, m/s^2\]

Step 6 ：Final Answer: The average acceleration over this time interval is \(\boxed{5 \, m/s^2}\).