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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) The acceleration was 2438 m/s². (b) The ball traveled 1.0971 meters.

Step by step solution

01

Convert Time Units

First, convert the contact time from milliseconds to seconds to make it compatible with other units. Since 1 millisecond is equal to \(0.001\) seconds, the contact time \( t \) is \( 30.0 \ \text{ms} = 30.0 \times 0.001 = 0.030 \ \text{s} \).
02

Calculate Acceleration

Using the formula for constant acceleration, \( v = u + at \), where \( v = 73.14 \, \text{m/s} \) is the final velocity, \( u = 0 \, \text{m/s} \) is the initial velocity, and \( t = 0.030 \, \text{s} \) is the time taken, we can solve for acceleration \( a \).\[ a = \frac{v - u}{t} = \frac{73.14 - 0}{0.030} = 2438 \, \text{m/s}^2 \]
03

Determine Distance Traveled

To find the distance \( s \), use the kinematic equation \( s = ut + \frac{1}{2}at^2 \). Substitute \( u = 0 \), \( a = 2438 \, \text{m/s}^2 \), and \( t = 0.030 \, \text{s} \) into the equation.\[ s = 0 \times 0.030 + \frac{1}{2} \times 2438 \times (0.030)^2 = \frac{1}{2} \times 2438 \times 0.0009 = 1.0971 \, \text{m} \]

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Key Concepts

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

constant acceleration
When discussing constant acceleration, it's essential to understand that this involves an object changing its velocity at a steady rate. This means that the net force acting upon the object is unchanging, leading to a consistent increase or decrease in speed over time. In the case of the tennis serve exercise, the tennis ball experiences constant acceleration while it is in contact with the racquet. Some everyday examples of constant acceleration include:
  • A car speeding up or slowing down on a straight highway.
  • A ball falling freely under gravity, where it accelerates at a constant rate of 9.81 m/s² downward.
During the serve, the ball starts from rest and gains speed steadily until it reaches its final velocity. The calculation involves using the formula for constant acceleration: \[ v = u + at \]where:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration, and
  • \( t \) is the time of contact.
The importance of consistent acceleration is key in predicting the ball's motion accurately.
problem solving in physics
Problem-solving in physics often requires a structured approach to break down complex situations into understandable steps. It involves analyzing given information, identifying which physics principles apply, selecting appropriate formulas, and carefully calculating to find unknown values. Let's take a look at how to approach such a problem using the tennis serve example:
  • Identify what is given and what is needed: You begin with knowing the contact time and the velocity at which the ball leaves the racquet. You need to find acceleration and distance traveled.
  • Convert units if necessary: Convert all measurements to compatible units. Here, the contact time was converted from milliseconds to seconds.
  • Choose the right formulas: Use appropriate equations, like the formula for acceleration and the kinematic equations, to find unknowns.
  • Check the reasonableness of your result: Once you have a result, it's important to verify whether it makes sense within the context of the problem.
Physics problems can become straightforward when you consistently apply these steps. Like any real-world scenario, this disciplined approach ensures that you solve problems systematically, making fewer mistakes and enhancing your understanding.
kinematic equations
Kinematic equations govern the motion of objects with constant acceleration. They are powerful tools in physics that connect the variables of motion, such as displacement, velocity, acceleration, and time.For uniformly accelerated motion, the primary kinematic equations are:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
These equations serve different purposes. For example, in our tennis serve problem, the first equation helped to calculate acceleration, knowing the initial and final velocities, and the time. The second equation allowed the determination of the distance the ball traveled while in contact with the racquet.To apply these equations:
  • Understand each variable: What data do they represent, and how are they physically connected?
  • Pick the right equation: Match the known variables with the unknowns you want to solve for.
  • Mind the units: Consistency is crucial to avoid calculation errors.
Mastering kinematic equations allows you not just to solve textbook problems but also to gain insights into everyday phenomena, like predicting how vehicles stop or objects fall.

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Most popular questions from this chapter

A world-class sprinter accelerates to his maximum speed in 4.0 \(\mathrm{s}\) . He then maintains this speed for the remainder of a \(100-\mathrm{m}\) race, finishing with a total time of 9.1 \(\mathrm{s}\) . (a) What is the runner's average acceleration during the first 4.0 \(\mathrm{s} ?(\mathrm{b})\) What is his average acceleration during the last 5.1 \(\mathrm{s} ?(\mathrm{c})\) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b).

An astronaut has left the Intemational Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a 10 -s interval. What are the magniude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the positive direction is to the right. (a) At the beginning of the interval the astronaut is moving toward the right along the \(x\) -axis at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end of the interval she is moving toward the right at 5.0 \(\mathrm{m} / \mathrm{s}\) . (b) At the beginning she is moving toward the left at \(5.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\) . (c) At the beginning she is moving toward the right at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\) .

You may have noticed while driving that your car's velocity does not continue to increase, even though you keep your foot on the gas pedal. This behavior is due to air resistance and friction between the moving parts of the car. Figure 2.48 shows a qualitative \(v_{x^{-}} t\) graph for a typical car if it starts from rest at the origin and travels in a straight line (the \(x\) -axis). Sketch qualitative \(a_{x}-t\) and \(x-t\) graphs for this car.

An apple drops from the tree and falls frecly. The apple is originally at rest a height \(H\) above the top of the grass of a thick lawn, which is made of blades of grass of height \(h\) . When the apple enters the grass, it slows down at a constant rate so that its speed is 0 when it reaches ground level. (a) Find the speed of the apple just before it enters the grass. (b) Find the acceleration of the apple while it is in the grass. (c) Sketch the \(y-t, v_{y}-t,\) and \(a_{y}-t\) graphs for the apple's motion.

The position of a particle between \(t=0\) and \(t=2.00 \mathrm{s}\) is given by \(x(t)=\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}-\left(10.0 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}+(9.00 \mathrm{m} / \mathrm{s}) t\) (a) Draw the \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs of this particle. (b) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle instantaneously at rest? Does your numerical result agree with the \(v_{x}-t\) graph in part \((a) ?\) At each time calculated in part \((b)\) is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from \(a_{x}(t)\) and from the \(v_{x}-t\) graph. (d) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the velocity of the particle instantancously not changing? Locate this point on the \(v_{x} t\) and \(a_{x}-t\) graphs of part (a). (e) What is the particle's greatest distance from the origin \((x=0)\) between \(t=0\) and \(t=2.00 \mathrm{s} ?(\mathrm{f})\) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle speeding \(u p\) at the greatest rate? At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle slowing down at the greatest rate? Locate these points on the \(v_{x}-t\) and \(a_{x}-t\) graphs of part (a).

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