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Chapter 2: Problem 27

A rubber ball is attached to a paddle by a rubber band. The ball is initially moving away from the paddle with a speed of \(4.0 \mathrm{m} / \mathrm{s} .\) After \(0.25 \mathrm{s}\), the ball is moving toward the paddle with a speed of \(3.0 \mathrm{m} / \mathrm{s} .\) What is the average acceleration of the ball during that 0.25 s? Give magnitude and direction.

Short Answer

Expert verified
Answer: The magnitude of the average acceleration is 28.0 m/s^2, and its direction is towards the paddle.

Step by step solution

01

Identify the given values

The initial velocity, \(v_i\), is given as \(4.0 \, \mathrm{m/s}\) away from the paddle. The final velocity, \(v_f\), is given as \(3.0 \, \mathrm{m/s}\) toward the paddle. The time interval is given as \(0.25 \, \mathrm{s}\).
02

Determine the final and initial velocities considering direction

Since the initial movement is away from the paddle, we can consider it as positive. Therefore, \(v_i = +4.0 \, \mathrm{m/s}\). The final velocity is towards the paddle, which is in the opposite direction. So, \(v_f = -3.0 \, \mathrm{m/s}\).
03

Calculate the average acceleration

Using the average acceleration formula: average acceleration \(= \frac{v_f - v_i}{t} = \frac{-3.0 \, \mathrm{m/s} - 4.0 \, \mathrm{m/s}}{0.25 \, \mathrm{s}} = \frac{-7.0 \, \mathrm{m/s}}{0.25 \, \mathrm{s}} = -28.0 \, \mathrm{m/s^2}\)
04

Determine the direction

Since the average acceleration is negative, it means the direction is opposite to the original movement (away from the paddle). Therefore, the direction of the average acceleration is towards the paddle. The magnitude of the average acceleration during the \(0.25 \, \mathrm{s}\) time interval is \(28.0 \, \mathrm{m/s^2}\), and its direction is towards the paddle.

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