Acceleration, 1D problem, chapter 2 basically

In summary, the problem involves a 50.0-g superball traveling at 25.0 m/s, bouncing off a brick wall and rebounding at 22.0 m/s. With a high-speed camera recording the event, the question asks for the magnitude of the average acceleration of the ball during the 3.50 ms (.0035 s) contact time with the wall. Using the equation \bar{a}=\frac {\Delta v}{\Delta t}=\frac {v_{f}-v_{i}}{t_{f}-t_{i}}, the solution is found to be 1.34 x 10^4 m/s^2. However, the answer in the book is given as 1
  • #1

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

[tex] \bar{a}= \frac {\Delta v}{\Delta t} = \frac {v_{f}-v_{i}}{t_{f}-t_{i}} [/tex]

The Attempt at a Solution

So the information from the problem gives me:

[tex] v_{i}=25 \frac{m}{s} [/tex]

[tex] v_{f} = 22 \frac {m}{s} [/tex]

[tex] t_{i} = 0 s [/tex]

[tex] t_{f} = .0035s [/tex]

so the average acceleration during the time interval should be:

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

only problem is that the book has 1.34 x 104 m/s^2 as the answer, what am I missing here?
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  • #2
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.
  • #3
The velocity vector has changed direction. What's the magnitude of the change in velocity?
  • #4
ah ok,


[tex] v_{f}=-22\frac{m}{s} [/tex]

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

By magnitude they do mean the absolute value of the result right?
  • #5
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.
  • #6
ok thanks for the help!

1. What is acceleration?

Acceleration is the rate of change of an object's velocity over time. In other words, it is the increase or decrease in an object's speed per unit time.

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in an object's velocity by the change in time. The formula for acceleration is: a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between positive and negative acceleration?

Positive acceleration occurs when an object's speed increases over time, while negative acceleration (also known as deceleration) occurs when an object's speed decreases over time. Both positive and negative acceleration are considered forms of acceleration.

4. Can an object have both constant speed and acceleration?

No, an object cannot have both constant speed and acceleration. If an object has a constant speed, then its acceleration is zero. This means that its velocity is not changing over time.

5. How does acceleration relate to the motion of an object in a 1D problem?

In a 1D problem, the motion of an object is described along a straight line. Acceleration affects the speed and direction of the object's motion along this line. It can cause the object to speed up, slow down, or change direction.

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