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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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Most popular questions from this chapter

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A stone is thrown vertically downward from the roof of a building. It passes a window \(16.0 \mathrm{m}\) below the roof with a speed of $25.0 \mathrm{m} / \mathrm{s} .\( It lands on the ground \)3.00 \mathrm{s}$ after it was thrown. What was (a) the initial velocity of the stone and (b) how tall is the building?
A relay race is run along a straight-line track of length \(300.0 \mathrm{m}\) running south to north. The first runner starts at the south end of the track and passes the baton to a teammate at the north end of the track. The second runner races back to the start line and passes the baton to a third runner who races \(100.0 \mathrm{m}\) northward to the finish line. The magnitudes of the average velocities of the first, second, and third runners during their parts of the race are \(7.30 \mathrm{m} / \mathrm{s}, 7.20 \mathrm{m} / \mathrm{s}\) and \(7.80 \mathrm{m} / \mathrm{s},\) respectively. What is the average velocity of the baton for the entire race? [Hint: You will need to find the time spent by each runner in completing her portion of the race.]
please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A penny is dropped from the observation deck of the Empire State building (369 m above ground). With what velocity does it strike the ground?
A train, traveling at a constant speed of \(22 \mathrm{m} / \mathrm{s}\), comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude $1.4 \mathrm{m} / \mathrm{s}^{2} .\( (a) Draw a graph of \)v_{x}\( versus \)t\( where the \)x$ -axis points up the incline. (b) What is the speed of the train after $8.0 \mathrm{s}$ on the incline? (c) How far has the train traveled up the incline after \(8.0 \mathrm{s} ?\) (d) Draw a motion diagram, showing the trains position at 2.0 -s intervals.
For the train of Example \(2.2,\) find the average velocity between 3: 14 P.M. when the train is at \(3 \mathrm{km}\) east of the origin and 3: 28 P.M. when it is \(10 \mathrm{km}\) east of the origin.
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