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Chapter 1: Problem 3

Suppose that an accelerating car goes from 0 mph to 61.4 mph in five seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: \(1 \mathrm{mph}\) is \(22 / 15 \mathrm{ft} / \mathrm{sec} .)\) Find the average acceleration of the car over each of the first two seconds. $$\begin{array}{|l|l|l|l|l|l|l|} \hline t(\mathrm{~s}) & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline v(t)(\mathrm{ft} / \mathrm{s}) & 0.00 & 30.68 & 53.18 & 69.55 & 81.82 & 90.00 \\ \hline \end{array}$$ average acceleration over the first second \(=\square\) help (units) average acceleration over the second second =\(\square\) help (units)

Short Answer

Expert verified
Average acceleration over first second is 30.68 ft/s\textsuperscript{2}, over the second second is 22.5 ft/s\textsuperscript{2}.

Step by step solution

01

Understand the Problem

The goal is to find the average acceleration of a car over specific time intervals. The velocity data is given at different time points.
02

Convert Given Information

Note that the velocity values are already converted from miles per hour to feet per second using the provided conversion factor (1 mph = 22/15 ft/s). No further conversion is needed.
03

Find Velocity Change for First Second

We need to determine the velocity change over the first second. From the table, at t=0 seconds, the velocity is 0 ft/s, and at t=1 second, the velocity is 30.68 ft/s. So, the change in velocity is 30.68 ft/s - 0 ft/s = 30.68 ft/s.
04

Calculate Average Acceleration for First Second

Average acceleration is given by the formula \(a = \frac{螖v}{螖t}\). Here, the change in time (螖t) is 1 second. So, average acceleration over the first second is \[a = \frac{30.68 \text{ ft/s}}{1 \text{ s}} = 30.68 \text{ ft/s}^2.\]
05

Find Velocity Change for Second Second

For the second second (from t=1 to t=2 seconds), the velocity at t=1 second is 30.68 ft/s, and at t=2 seconds, the velocity is 53.18 ft/s. So, the change in velocity is 53.18 ft/s - 30.68 ft/s = 22.5 ft/s.
06

Calculate Average Acceleration for Second Second

Using the same formula \(a = \frac{螖v}{螖t}\), the change in time (螖t) is 1 second here as well. So, the average acceleration over the second second is \[a = \frac{22.5 \text{ ft/s}}{1 \text{ s}} = 22.5 \text{ ft/s}^2.\]

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Key Concepts

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

Velocity Change
Velocity change refers to the difference in velocity over a specific period. When dealing with acceleration, understanding how the velocity of an object varies over time is crucial. In our exercise, we dealt with a car accelerating over seconds. Velocity is provided at different points:
  • At 0 seconds: 0 ft/s
  • At 1 second: 30.68 ft/s
  • At 2 seconds: 53.18 ft/s
For the first second, the velocity changed from 0 ft/s to 30.68 ft/s, giving a velocity change of 30.68 ft/s.

For the second second, it changed from 30.68 ft/s to 53.18 ft/s, resulting in a velocity change of 22.5 ft/s.

Understanding how to calculate the change in velocity is essential in determining acceleration.
Time Intervals
The concept of time intervals is closely linked to determining acceleration. Time intervals refer to the periods over which we are measuring changes, such as changes in velocity.

In the context of our example, we looked at two main time intervals:
  • From 0 to 1 second (first second)
  • From 1 to 2 seconds (second second)
Each time interval represents a one-second period during which we measured the car's velocity. By focusing on these intervals, we can correctly assess the changes in velocity specific to each period.

Understanding these time intervals ensures we correctly apply the acceleration formula.
Acceleration Formula
Acceleration is a measure of how quickly velocity changes over time. To find average acceleration, we use the formula:

\( a = \frac{螖v}{螖t} \)

Where:
\(螖v \) = change in velocity
\(螖t \) = change in time

In our example, we had two intervals:

First Second:
\(螖v = \) 30.68 ft/s (velocity change)
\(螖t = \) 1 second (time interval)
\( a = \frac{30.68 \text{ ft/s}}{1 \text{ s}} = 30.68 \text{ ft/s}^2 \)

Second Second:
\(螖v = \) 22.5 ft/s
\(螖t = \) 1 second
\( a = \frac{22.5 \text{ ft/s}}{1 \text{ s}} = 22.5 \text{ ft/s}^2 \)

Applying the acceleration formula in each time interval allowed us to find the car's average acceleration during those intervals. Always ensure to identify the correct velocity and time changes before using this formula.

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Most popular questions from this chapter

Consider a car whose position, \(s\), is given by the table $$\begin{array}{|l|l|l|l|l|l|l|} \hline t(\mathrm{~s}) & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 \\ \hline s(\mathrm{ft}) & 0 & 0.45 & 1.7 & 3.8 & 6.5 & 9.6 \\ \hline \end{array}$$ Find the average velocity over the interval \(0 \leq t \leq 0.2\). average velocity \(=\square\) help (units) Estimate the velocity at \(t=0.2\) velocity \(=\square\) help (units)

The temperature, \(H\), in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by \(H=f(t),\) where \(t\) is in minutes since the coffee was put on the counter. (a) Is \(f^{\prime}(t)\) positive or negative? [Choose: positive | negative] (Be sure that you are able to give a reason for your answer.) (b) What are the units of \(f^{\prime}(35)\) ?\(\square\) heln (units) Suppose that \(\left|f^{\prime}(35)\right|=1.2\) and \(f(35)=52 .\) Fill in the blanks (including units where needed) and select the appropriate terms to complete the following statement about the temperature of the coffee in this case. At \(\square\) minutes after the coffee was put on the counter, its [Choose: derivative | temperature change in temperature] is \(\square\) and will [Choose: increase | decrease] by about \(\square\) in the next 90 seconds.

A bungee jumper dives from a tower at time \(t=0\). Her height \(h\) (measured in feet) at time \(t\) (in seconds) is given by the graph in Figure 1.1.4. In this problem, you may base your answers on estimates from the graph or use the fact that the jumper's height function is given by \(s(t)=100 \cos (0.75 t) \cdot e^{-0.2 t}+100\). a. What is the change in vertical position of the bungee jumper between \(t=0\) and \(t=15 ?\) b. Estimate the jumper's average velocity on each of the following time intervals: [0,15] , \([0,2],[1,6],\) and \([8,10] .\) Include units on your answers. c. On what time interval(s) do you think the bungee jumper achieves her greatest average velocity? Why? d. Estimate the jumper's instantaneous velocity at \(t=5\). Show your work and explain your reasoning, and include units on your answer. e. Among the average and instantaneous velocities you computed in earlier questions, which are positive and which are negative? What does negative velocity indicate?

The temperature change \(T\) (in Fahrenheit degrees), in a patient, that is generated by a dose \(q\) (in milliliters), of a drug, is given by the function \(T=f(q)\). a. What does it mean to say \(f(50)=0.75 ?\) Write a complete sentence to explain, using correct units. b. A person's sensitivity, \(s,\) to the drug is defined by the function \(s(q)=f^{\prime}(q) .\) What are the units of sensitivity? c. Suppose that \(f^{\prime}(50)=-0.02 .\) Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a).

According to the U.S. census, the population of the city of Grand Rapids, MI, was 181,843 in \(1980 ; 189,126\) in 1990 ; and 197,800 in 2000 . a. Between 1980 and 2000 , by how many people did the population of Grand Rapids grow? b. In an average year between 1980 and 2000 , by how many people did the population of Grand Rapids grow? c. Just like we can find the average velocity of a moving body by computing change in position over change in time, we can compute the average rate of change of any function \(f\). In particular, the average rate of change of a function \(f\) over an interval \([a, b]\) is the quotient $$\frac{f(b)-f(a)}{b-a}$$ What does the quantity \(\frac{f(b)-f(a)}{b-a}\) measure on the graph of \(y=f(x)\) over the interval \([a, b] ?\) d. Let \(P(t)\) represent the population of Grand Rapids at time \(t,\) where \(t\) is measured in years from January \(1,1980 .\) What is the average rate of change of \(P\) on the interval \(t=0\) to \(t=20 ?\) What are the units on this quantity? e. If we assume the population of Grand Rapids is growing at a rate of approximately \(4 \%\) per decade, we can model the population function with the formula $$P(t)=181843(1.04)^{t / 10}$$ Use this formula to compute the average rate of change of the population on the intervals \([5,10],[5,9],[5,8],[5,7],\) and [5,6] f. How fast do you think the population of Grand Rapids was changing on January 1 , 1985 ? Said differently, at what rate do you think people were being added to the population of Grand Rapids as of January \(1,1985 ?\) How many additional people should the city have expected in the following year? Why?
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