Question adapted from Examkrackers MCAT prep book:
A particle moves along a half circle (diameter=$10\text{ m}$) at a constant speed of $1\text{ m/s}$. What is the average acceleration of the particle as it moves from one side of the half circle to the other side?
A. $0$
B. $0.2/\pi$
C. $0.4/\pi$
D. $1$
The book says C is correct. Acceleration is change in velocity divided by time. Initial velocity is $1\text{ m/s}$ up; final velocity is $1\text{ m/s}$ down. The change in velocity is therefore $2\text{ m/s}$. The time is found from speed equals distance divided by time. Distance is $2\pi r/2$. Thus
$$a= \frac{2}{(2\pi(5)/2)/1} = \frac{2}{5\pi} = 0.4/\pi$$
I thought all of the answers were wrong because I thought they should've used the centripetal acceleration equation: $a= v^2/r$
SO my question like the title: Is there a difference between "average acceleration" and centripetal acceleration?
I searched for a couple hours and couldn't find this issue directly addressed.