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A Tennis Serve. In the fastest measured tennis serve, the ball left the racquet at 73.14 \(\mathrm{m} / \mathrm{s} .\) A served tennis ball is typically in contact with the racquet for 30.0 \(\mathrm{ms}\) and starts from rest. Assume constant acceleration. (a) What was the ball's acceleration during this serve? (b) How far did the ball travel during the serve?

Short Answer

Expert verified
(a) 2438 m/s² (b) 1.0971 m

Step by step solution

01

Identify known values and convert units

We are given that the final velocity of the tennis ball is \( v_f = 73.14 \, \text{m/s} \) and the initial velocity \( v_i = 0 \, \text{m/s} \) since the ball starts from rest. The time of contact with the racquet is \( t = 30.0 \, \text{ms} = 0.030 \, \text{s} \).
02

Use the formula for constant acceleration

To find acceleration, we use the formula for constant acceleration:\[ a = \frac{v_f - v_i}{t} \]Substitute the known values:\[ a = \frac{73.14 \, \text{m/s} - 0 \, \text{m/s}}{0.030 \, \text{s}} \]
03

Calculate the acceleration

Calculate the acceleration using the values from Step 2:\[ a = \frac{73.14}{0.030} = 2438 \text{ m/s}^2 \]
04

Use the formula for distance traveled under constant acceleration

We use the formula:\[ s = v_i t + \frac{1}{2} a t^2 \]Since \( v_i = 0 \), the formula simplifies to:\[ s = \frac{1}{2} a t^2 \]
05

Calculate the distance traveled

Substitute the known values into the simplified distance formula:\[ s = \frac{1}{2} \times 2438 \, \text{m/s}^2 \times (0.030 \, \text{s})^2 \]Calculate the distance:\[ s = \frac{1}{2} \times 2438 \times 0.0009 = 1.0971 \, \text{m} \]
06

Conclusion: Answer to the parts of the problem

(a) The ball's acceleration during the serve is \( 2438 \, \text{m/s}^2 \).(b) The ball traveled \( 1.0971 \, \text{m} \) during the serve.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing the motion. When dealing with problems such as a tennis serve, we often use kinematics to predict variables like velocity, position, and acceleration over a given time.
  • Kinematics focuses on measurable quantities: displacement, velocity, and acceleration.
  • It provides a mathematical approach to describe motion and to predict the future position and velocity of an object.
  • Key equations in kinematics are derived from calculus and based on assumptions such as constant acceleration.
In our tennis serve example, we have essential kinematical parameters: initial velocity of 0 m/s (as the ball starts from rest), a known final velocity, and the contact time between the ball and the racquet. These parameters allow us to apply kinematic equations to figure out the ball’s acceleration and the distance it travels during the serve.
Exploring Velocity in Motion
Velocity is a vector quantity that refers to the rate at which an object changes its position. It is an essential part of kinematic equations, especially when dealing with constant acceleration.
  • Velocity can be defined as the change in position with respect to time and has both magnitude (speed) and direction.
  • Unlike speed, velocity can be negative, indicating the direction of motion.
  • When acceleration is constant, average velocity can be calculated as the sum of the initial and final velocities divided by two.
In the tennis serve problem, we know that the ball starts from rest and reaches a velocity of 73.14 m/s. Using this information, and knowing the time of contact (0.030 s), we can calculate the acceleration, which in turn allows us to explore other aspects of the ball’s motion.
Calculating Distance Traveled
Distance calculation in kinematics often involves integrating velocity or using time, initial velocity, acceleration, and other known values to determine how far an object travels.
  • Distance traveled, or displacement, can be calculated using the formula: \( s = v_i t + \frac{1}{2} a t^2 \).
  • In this formula, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
  • When the initial velocity is zero, the formula simplifies because the \( v_i t \) term becomes zero.
In our scenario, since the ball begins at rest, we can directly use the simplified formula \( s = \frac{1}{2} a t^2 \) to calculate the distance. Substituting the calculated acceleration and the given time period, we determine the ball traveled approximately 1.0971 meters during this brief encounter with the racquet. Understanding these calculations provides deeper insights into the world of motion and helps illustrate how predictions about distance and speed can be made with precision using kinematic principles.

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Most popular questions from this chapter

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s} .\) Air resistance may be ignored. (a) At what time after being ejected is the boulder moving at 20.0 \(\mathrm{m} / \mathrm{s}\) upward? (b) At what time is it moving at 20.0 \(\mathrm{m} / \mathrm{s}\) downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-\) t graphs for the motion.

You throw a glob of putty straight up toward the ceiling, which is 3.60 \(\mathrm{m}\) above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 \(\mathrm{m} / \mathrm{s}\). (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling?

A bird is flying due east. Its distance from a tall building is given by \(x(t)=28.0 \mathrm{m}+(12.4 \mathrm{m} / \mathrm{s}) t-\left(0.0450 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) What is the instantaneous velocity of the bird when \(t=8.00 \mathrm{s}?\)

Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s} .\) If you start 0.90 m behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 \(\mathrm{m}\) away at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) , returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?

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