Is there a difference between “average acceleration” and centripetal acceleration?

Question

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.

Answer 1

Is there a difference between "average acceleration" and centripetal acceleration?

Yes, in fact they're almost completely unrelated.

The average acceleration is defined as

$$\vec a_\text{avg} = \frac{\Delta\vec v}{\Delta t}$$

It is one quantity that partially describes the motion of a particle over an extended time. In other words, average acceleration encapsulates the fact that a particle started with some velocity at time A and ended with some velocity at time B, but completely ignores what the particle did between A and B. This is by design.

Centripetal acceleration, on the other hand, is an instantaneous quantity: it's the radial component of acceleration. (This requires that you have chosen some point to be the center of a polar coordinate system.) It partially describes the motion of a particle at one moment, not over an extended time.

Answer 2

Here the average acceleration can be understood as follows:

The particle going from A to B along a half circle with speed 1m/s can be here viewed as the particle going from A to C and again to A with an initial velocity 1m/s as shown in the figure:

The particle in the half circle, will move under a centripetal acceleration which is always directed towards the center. On the other hand the average acceleration is directed downwards, as it is the final velocity(downward) minus initial velocity(upward) divided by time(vector difference along a line). That is why considering only the y-component of the velocity of the particle in the half circle path, the half circle path is reduced to a straight path for the average acceleration.

As we see in the second figure, there is an acceleration in the downward direction that decelerates the particle when it goes from A to C and accelerate it when it comes to A again. This constant acceleration in the straight path is the average acceleration for the actual half circle path. That is why the average acceleration is calculated as the difference of initial and final velocity divided by time, difference of up velocity and down velocity at A divided by time.

Answer 3

The problem with centripetal acceleration is that it is not a vector, and cannot possible have a negative sign. It should remain "constant" in this case, but its direction is changing. But actually, it's not remaining constant, just the magnitude is remaining constant while the direction is changing. Centripetal acceleration will just give you the magnitude of the acceleration a direction towards the center of the circle.

Answer 4

Centripetal acceleration is a type of acceleration but Average acceleration is the calculation of acceleration. They are different.