Acceleration, 1D problem, chapter 2 basically

[Asphyxiated]

Homework Statement

A 50.0-g superball is traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms (.0035 s) what is the magnitude of the average acceleration of the ball during this time interval?

Homework Equations

\bar{a}= \frac {\Delta v}{\Delta t} = \frac {v_{f}-v_{i}}{t_{f}-t_{i}}

The Attempt at a Solution

So the information from the problem gives me:

v_{i}=25 \frac{m}{s}

v_{f} = 22 \frac {m}{s}

t_{i} = 0 s

t_{f} = .0035s

so the average acceleration during the time interval should be:

\bar{a} = \frac {22-25}{.0035-0}=|-\frac {3}{.0035}| \approx 857.14 \frac {m}{s^{2}}

only problem is that the book has 1.34 x 10⁴ m/s^2 as the answer, what am I missing here?

 

[rl.bhat]

In given problem, the directions of initial and final velocities are not the same. So find the change in velocities taking into consideration the directions.

 

[gneill]

The velocity vector has changed direction. What's the magnitude of the change in velocity?

 

[Asphyxiated]

ah ok,

so

v_{f}=-22\frac{m}{s}

\bar{a}=\frac{-22-25}{.0035}=|\frac{-47}{.0035}| \approx 13428.6 \frac {m}{s^{2}} = 1.34*10^{4}\frac{m}{s^{2}}

By magnitude they do mean the absolute value of the result right?

 

[gneill]

Yes, they want the magnitude of the average acceleration vector, that is, its "size" without its direction. That's why I suggested taking the magnitude of the velocity change. You could have found the velocity change and acceleration first (yielding a negative value) and then taken its magnitude after.

 

[Asphyxiated]

ok thanks for the help!