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Chapter 1: Problem 7

The velocity of a ball that has been tossed vertically in the air is given by \(v(t)=16-32 t\), where \(v\) is measured in feet per second, and \(t\) is measured in seconds. The ball is in the air from \(t=0\) until \(t=2\). a. When is the ball's velocity greatest? b. Determine the value of \(v^{\prime}(1)\). Justify your thinking. c. What are the units on the value of \(v^{\prime}(1) ?\) What does this value and the corresponding units tell you about the behavior of the ball at time \(t=1 ?\) d. What is the physical meaning of the function \(v^{\prime}(t) ?\)

Short Answer

Expert verified
a. Greatest at \(t = 0\) with \(v(0) = 16 \text{ft/s}\). b. \(v^{\prime}(1) = -32 \text{ft/s}^2.\) c. The units are ft/s^2, indicating deceleration at 32 ft/s^2 at t=1. d. \(v^{\prime}(t)\) is the ball's acceleration.

Step by step solution

01

Determine the time interval

The ball is in the air from \(t = 0\) to \(t = 2\) seconds. This will help to find the range for velocity calculation.
02

Find when velocity is greatest

The velocity function is v(t) = 16 - 32t. It is a linear function. The velocity is greatest at the smallest value of t. Hence, v(0) is the greatest velocity. Calculate v(0): \[v(0) = 16 - 32 \times 0 = 16 \text{ feet per second} \]
03

Calculate the derivative of v(t)

To determine \(v^{\prime}(t)\), find the derivative of \(v(t) = 16 - 32t\). \[v^{\prime}(t) = -32 \]
04

Determine the value of \(v^{\prime}(1)\)

Substitute \(t = 1\) into \(v^{\prime}(t)\): \[v^{\prime}(1) = -32 \text{ feet per second squared} \]
05

Explain the units of \(v^{\prime}(1)\)

The units of \(v^{\prime}(1)\) are feet per second squared (ft/s^2). This tells us the rate of change of velocity with respect to time.
06

Interpret \(v^{\prime}(1)\)

At \(t = 1\), the velocity of the ball is decreasing by 32 feet per second every second. This means the ball is decelerating at 32 feet per second squared.
07

Determine the physical meaning of \(v^{\prime}(t)\)

\(v^{\prime}(t)\) represents the acceleration of the ball, which is the rate of change of velocity with respect to time.

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Key Concepts

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

Understanding the Derivative in Calculus
The derivative is a fundamental concept in calculus. It represents how a function changes as its input changes. For any function, the derivative tells us the rate at which the function is increasing or decreasing. In this exercise, we looked at the velocity of a ball given by the function \(v(t) = 16 - 32t\).
To find how this velocity changes over time, we took the derivative of the velocity function. Mathematically, we calculated \(v^{\text{'} }(t) = -32\). This gives us the rate of change of velocity with respect to time.
This process of finding the derivative is crucial in physics and many other fields to understand how various quantities change.
What is Acceleration?
Acceleration is a key concept in physics that refers to how quickly an object's velocity changes. It is the derivative of the velocity function. From the exercise, we determined that \(v^{\text{'}}(t) = -32\). This tells us that the acceleration of the ball is constant at \( -32 \text{ ft/s}^2 \).
Negative acceleration indicates that the ball is slowing down; hence, it's often called deceleration.
Understanding acceleration helps us predict how the velocity of the ball changes over time.
The Concept of Rate of Change
The rate of change measures how a quantity changes with respect to another quantity. In calculus, this concept is embodied by the derivative. In the context of the exercise, the rate of change of the velocity function \(v(t) = 16 - 32t\) is represented by the derivative \(v^{\text{'}}(t) = -32\).
This constant rate of change reveals how the velocity of the ball changes every second. Specifically, as each second passes, the velocity decreases by 32 feet per second.
Rates of change are pivotal in understanding how different quantities evolve over time.
Linear Functions and Their Characteristics
A linear function is one in the form \(f(t) = mt + b\), where \(m\) and \(b\) are constants. The velocity function \(v(t) = 16 - 32t\) is a linear function with a slope of -32 and y-intercept of 16.
Linear functions have a constant rate of change, which is reflected in their straight-line graphs. In our case, the slope -32 tells us how steeply the velocity changes.
The y-intercept 16 indicates the initial velocity of the ball when \(t = 0\). Linear functions are straightforward and predictable, making them easy to analyze in physics problems.
Importance of Units in Physics
Units are fundamental in physics because they provide a standard for measurement. In this exercise, velocity is given in feet per second (ft/s), and acceleration in feet per second squared (ft/s虏).
Understanding the units helps interpret the physical meaning of calculated values. For instance, \(v^{\text{'}}(1) = -32 \text{ feet per second squared}\) tells us the rate at which the ball's velocity is decreasing each second.
Consistency in units ensures that our calculations are meaningful and comparable. Never underestimate the need for clear units in any physics-related problem.

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Most popular questions from this chapter

For each of the following prompts, give an example of a function that satisfies the stated criteria. A formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why. a. A function \(f\) that is continuous at \(a=2\) but not differentiable at \(a=2\). b. A function \(g\) that is differentiable at \(a=3\) but does not have a limit at \(a=3\). c. A function \(h\) that has a limit at \(a=-2,\) is defined at \(a=-2,\) but is not continuous at \(a=-2\) d. A function \(p\) that satisfies all of the following: \- \(p(-1)=3\) and \(\lim _{x \rightarrow-1} p(x)=2\) $$\text { - } p(0)=1 \text { and } p^{\prime}(0)=0$$ \- \(\lim _{x \rightarrow 1} p(x)=p(1)\) and \(p^{\prime}(1)\) does not exist

Suppose that the population, \(P\), of China (in billions) can be approximated by the function \(P(t)=1.15(1.014)^{t}\) where \(t\) is the number of years since the start of \(1993 .\) a. According to the model, what was the total change in the population of China between January 1,1993 and January \(1,2000 ?\) What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on January \(1,2000 ?\) Explain and justify, being sure to include proper units on all your answers. b. According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, \(2012 ?\) c. Write an expression involving limits that, if evaluated, would give the exact instantaneous rate of change of the population on today's date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (including units) of the value you have found. d. Find an equation for the tangent line to the function \(y=P(t)\) at the point where the \(t\) -value is given by today's date.

The cost, \(C\) (in dollars) to produce \(g\) gallons of ice cream can be expressed as \(C=f(g)\). (a) In the expression \(f(300)=350\), what are the units of \(300 ?\) [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] what are the units of \(350 ?\) [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] (b) In the expression \(f^{\prime}(300)=1.2,\) what are the units of 300? [Choose:? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] what are the units of 1.2? [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] (Be sure that you can carefully put into words the meanings of each of these statement in terms of ice cream and money.)

Consider the function whose formula is \(f(x)=\frac{16-x^{4}}{x^{2}-4}\). a. What is the domain of \(f ?\) b. Use a sequence of values of \(x\) near \(a=2\) to estimate the value of \(\lim _{x \rightarrow 2} f(x)\), if you think the limit exists. If you think the limit doesn't exist, explain why. c. Use algebra to simplify the expression \(\frac{16-x^{4}}{x^{2}-4}\) and hence work to evaluate \(\lim _{x \rightarrow 2} f(x)\) exactly, if it exists, or to explain how your work shows the limit fails to exist. Discuss how your findings compare to your results in (b). d. True or false: \(f(2)=-8\). Why? e. True or false: \(\frac{16-x^{4}}{x^{2}-4}=-4-x^{2}\). Why? How is this equality connected to your work above with the function \(f ?\) f. Based on all of your work above, construct an accurate, labeled graph of \(y=f(x)\) on the interval \([1,3],\) and write a sentence that explains what you now know about \(\lim _{x \rightarrow 2} \frac{16-x^{4}}{x^{2}-4}\)

For a certain function \(f\), its derivative is known to be \(f^{\prime}(x)=(x-1) e^{-x^{2}}\). Note that you do not know a formula for \(y=f(x)\). a. At what \(x\) -value(s) is \(f^{\prime}(x)=0\) ? Justify your answer algebraically, but include a graph of \(f^{\prime}\) to support your conclusion. b. Reasoning graphically, for what intervals of \(x\) -values is \(f^{\prime \prime}(x)>0\) ? What does this tell you about the behavior of the original function \(f ?\) Explain. c. Assuming that \(f(2)=-3\), estimate the value of \(f(1.88)\) by finding and using the tangent line approximation to \(f\) at \(x=2\). Is your estimate larger or smaller than the true value of \(f(1.88) ?\) Justify your answer.
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