# Average acceleration

1. Feb 19, 2008

### chocolatelover

1. The problem statement, all variables and given/known data
A 45.0 g Super Ball traveling at 30.0 m/s bounces off a brick wall and rebounds at 21m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.0ms, what is the magnitude of the average acceleration of the ball during this time interval? (It should be positive)

2. Relevant equations
1ms=10^-3s
average acceleration=final velocity-initial velocity/time interval

3. The attempt at a solution

21-30/3-0=-3ms

1ms(10^-3s/-3ms)=-.00033

but this can't be right because it's negative, right?

Thank you very much

Last edited: Feb 20, 2008
2. Feb 19, 2008

### PhanthomJay

The negative is OK, but your magnitude is incorrect. You must watch your plus and minus signs in your designations of the initial and final velocities...they are in different directions. Also, if you end up with a negative for the acceleration, you should interpret its meaning.

3. Feb 19, 2008

### brendan3eb

I am not sure if you have gotten to impulse, but if you have, then you can solve with those equations. Recall that linear momentum J equals the change in impulse - J=mv2-mv1

Also remember that J=Favg*T
You'll have solved for J, and you already know that T=3 ms, so you should be able to find your average force during this period of time. Your average force should equal mass times your average acceleration.
F=ma
a=Favg/m

That's how I would solve it. Let me know if that helps.

4. Feb 19, 2008

### chocolatelover

Thank you very much

5. Feb 20, 2008

### chocolatelover

Does this look correct?

I know that the average acceleration=velocity/time

So couldn't I do the following?

Average acceleration=-21-30/3-0
= -17

In order to find the magnitude, I would need to put this in scientific notation, right?

-1.7 X 10^1

The magnitude would be 1, right?

Thank you very much

6. Feb 20, 2008

### PhanthomJay

You have got to be consistent with your units. The average acceleration is the change in velocity over the change in time. Velocity is given in meters per second; you should therfore first convert 3 milliseconds to seconds, before doing the division. What will be the units of the acceleration?

Magnitude means 'how much', it does not refer to the scientific notation exponent; it just means "how much is the acceleration?" You don't necessarily have to convert to scientific notation, although it sometimes makes the result clearer by avoiding a lot of zeros.

7. Feb 20, 2008

### chocolatelover

Thank you very much

Does this look right?

-21-30/.003-0= -17000

1000ms=1s
3ms=.003

But it says that it should be positive

Thank you

8. Feb 20, 2008

### PhanthomJay

math is good, but what are the units of the acceleration? The magnitude of the acceleration is 17,000 (____) (fill in the blank with the correct units; if the velocity is in m/s and the time is in seconds, what are the units of the acceleration?).

9. Feb 20, 2008

### chocolatelover

Thank you very much

Regards

10. Feb 20, 2008

### PhanthomJay

You are welcome. Oh, by the way, you got a negative value because you assumed the initial velocity of the ball was positive to the right, and its final velocity was negative to the left, thus it's change in velocity was Vf-Vi= -21-30= -51m/s , where the minus sign means that the velocity change, and hence the acceleration, is to the left. The magnitude of the acceleration is just the absolute value ,or a positive number, of 17000m/s^2. The negative indicates the direction.