# Conceptual issue with average acceleration problem in Giancoli

1. Jul 22, 2014

### knyaz

Hi all,

I recently came upon the following problem in Giancoli's Physics Principles With Applications, 6th ed. textbook that I am having some issues understanding. I am new here, so I do apologize if I have made any mistakes in formatting or anything else.

1. The problem statement, all variables and given/known data

A 0.140-kg baseball traveling 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove?

2. Relevant equations

1.) favg = m*aavg

This is where I am struggling—the solutions manual to this book states that the following equation is also to be used:

2.) aavg = (v2 - v0 2) / 2(x - x0)

3. The attempt at a solution

I am familiar with that second equation up there when used to find constant acceleration, but I do not see how it can be used to find average acceleration. The derivation of this equation from the basic definitions of average velocity and acceleration involves using the following:

vavg = (v + v0)/2

As far as I can tell, this wouldn't work for average acceleration—at non-constant acceleration, the average velocity would not necessarily be halfway between the initial and final velocities. Thus, it seems to me that the equation the solutions manual uses, no.2 above, cannot be used in solving the problem.
Is the solutions manual wrong, then, and is the problem unsolvable (to anyone this far in the textbook, at any rate) as it is stated? Please do tell me if I am missing something in my reasoning on this, or if I am otherwise incorrect.
I also do want to mention that I saw a similar post on here, involving this exact conceptual issue with a very similar problem. However, there the problem had asked for an estimate, which meant that eq (2) above was alright to use as an estimate of average acceleration. Here, the problem seems to ask for an exact answer—I do not know if it was meant to ask for an estimate, instead.

Thanks for hearing me out on this, and thanks in advance to anyone who replies!

-Alex

2. Jul 22, 2014

### Staff: Mentor

Hi knyaz,

Welcome to Physics Forums.

I agree with you that there is some ambiguity here. If you combine equations 1 and 2 to eliminate the average acceleration, you get the correct result for the average force, if, by the average force, you mean the force averaged over the distance. And, you get the average acceleration by dividing the average force by the mass. So, the average acceleration here is the acceleration averaged over the distance.

But, if, by the average force, you mean the force averaged over the time, then no.

Chet

3. Jul 23, 2014

### haruspex

Hi knyaz,
I agree with you completely. My position is a little different from Chet's. I don't see "average force" as ambiguous. It means an average over time. If the question wants you to take an average over distance then it should specify that. This issue comes up regularly on this forum.

4. Jul 23, 2014

### Staff: Mentor

Hi guys,

I don't have as much of an issue working with two different (and conflicting) definitions of average force. One nice feature of the force averaged over the displacement is that it is related to the amount of work done.

But, if someone refers to the average force (without any qualification), I'm automatically thinking "force averaged over time." If the problem statement wants you to be working with the force averaged over the displacement, in my opinion, they must specify this explicitly. So, in short, I agree with you guys.

Chet

5. Jul 23, 2014

### BiGyElLoWhAt

I don't know if you're this far yet, but it would be really easy to solve this using energies. (Hinted at by chet)

6. Jul 23, 2014

### vela

Staff Emeritus
I concur with you and the others. The given solution at best yields an estimate of the average force.

The average acceleration is defined as $a_\text{avg} = \frac{\Delta v}{\Delta t}$. You can think of it as the constant acceleration that produces the same $\Delta v$ in the same $\Delta t$. So far so good. You're given $\Delta x$ and $\Delta v$. You need $\Delta t$. The question is, can you infer what $\Delta t$ is from $\Delta x$? If you assume the acceleration is constant, like the solutions did, you can use the familiar kinematic equations to essentially find $\Delta t$ and the resulting average acceleration.

The problem is that that assumption is bad. For different accelerations, the ball will travel different distances coming to rest. You can see this in the plots of velocity vs. time I've attached. For all curves, $\Delta v$ and $\Delta t$ are the same so the average accelerations are the same, but the displacement, $\Delta x$, which is represented by the area under the curves, is different for each curve. So the same average acceleration can correspond to different displacements. If you require that the displacement be 11.0 cm, then something else has to give, namely $\Delta t$. The time the ball takes to come to rest will depend on how the acceleration varies over time, resulting in different average accelerations.

#### Attached Files:

• ###### vt.png
File size:
5.3 KB
Views:
51
Last edited: Jul 23, 2014
7. Jul 23, 2014

### knyaz

Thanks everyone for the replies, I really do appreciate it.
It was mainly the means of obtaining the average acceleration given in the solution that I was having issues with, because as vela pointed out there are any number of ways in which the ball could come to rest (thanks vela, I actually understand this a lot better now). It was particularly confusing that the author had explicitly stated that that equation could only be used under the assumption that acceleration is constant, which is not necessarily the case here.

8. Jul 24, 2014

### haruspex

In many cases, SHM would be a better guess for an approximation of the impact profile. (It's interesting to work out the ratio between average over time and average over distance in that case.)
For the ball catcher scenario, I would think there is an initial very large force as the hand/arm is accelerated rapidly up to the speed of the ball, then a more-or-less constant force as the muscles bring the arm back to rest.