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Chapter 4: Problem 5

Suppose that an accelerating car goes from 0 mph to 66.8 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 \(22 / 15\) feet per sec \(=22 / 15 \mathrm{ft} / \mathrm{s}\).) Find the average acceleration of the car over each of the first two seconds. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline v(t) & 0.00 & 33.41 & 57.91 & 75.73 & 89.09 & 98.00 \\ \hline \end{array} $$ average acceleration over the first second \(=\) ______________ (include units) average aceleration over the second second = _______________ (include units)

Short Answer

Expert verified
Average acceleration for first second: 33.41 ft/s虏, for second second: 24.50 ft/s虏

Step by step solution

01

- Understand the given data

The table provides the velocity of the car at different time points. The goal is to find the average acceleration over each of the first two seconds. Velocity units are in feet per second, and time is in seconds.
02

- Average acceleration formula

The average acceleration can be calculated using the formula: \[a_{avg} = \frac{v_{f} - v_{i}}{t_{f} - t_{i}}\] where \(v_{f}\) and \(v_{i}\) are the final and initial velocities respectively, and \(t_{f}\) and \(t_{i}\) are the final and initial times.
03

- Calculate average acceleration for the first second

Using the velocities and times from the table for the first second: \[a_{avg, 1st} = \frac{v(1) - v(0)}{1 - 0} = \frac{33.41 - 0.00}{1 - 0} = 33.41 \text{ ft/s}^2\]
04

- Calculate average acceleration for the second second

Using the velocities and times from the table for the second second: \[a_{avg, 2nd} = \frac{v(2) - v(1)}{2 - 1} = \frac{57.91 - 33.41}{2 - 1} = 24.50 \text{ ft/s}^2\]

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Key Concepts

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

velocity table
Velocity tables are useful tools in physics to record and analyze the speed of an object at various time intervals. Here, it shows how fast the car is going at specific moments. The data usually includes time in seconds and velocity in feet per second (ft/s). For example, at time 0 seconds, the car's velocity is 0 ft/s, and at time 1 second, it's 33.41 ft/s. By looking at these values, you can see how the speed of an object changes over time. This is important for calculating average accelerations.
average acceleration formula
The average acceleration formula is crucial for understanding how an object's speed changes over a period. The formula is written as \[a_{avg} = \frac{v_{f} - v_{i}}{t_{f} - t_{i}}\] Here, \(v_{f}\) is the final velocity, \(v_{i}\) is the initial velocity, \(t_{f}\) is the final time, and \(t_{i}\) is the initial time. In simpler terms, average acceleration measures how much the velocity changes per unit of time. For example, if an object's speed increases from 0 ft/s to 33.41 ft/s in 1 second, the average acceleration will be \[ \frac{33.41 - 0.00}{1 - 0} = 33.41 \text{ ft/s}^2\] It's important to use consistent units (like ft/s) and the actual change in time for accurate results.
time intervals
Understanding time intervals is key to calculating average acceleration. A time interval is simply the difference between two time points. For example, the interval between 0 seconds and 1 second is 1 second. Knowing these intervals allows you to apply the average acceleration formula correctly.
  • For the first second: start time (t鈧�) = 0 s, end time (t鈧�) = 1 s
  • For the second second: start time (t鈧�) = 1 s, end time (t鈧�) = 2 s
To find the average acceleration for these intervals, you use the velocities at these specific times and plug them into the formula. This practice helps in understanding how speed changes over any given period.

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Most popular questions from this chapter

In Chapter \(1,\) we showed that for an object moving along a straight line with position function \(s(t),\) the object's "average velocity on the interval \([a, b]^{\prime \prime}\) is given by $$ A V_{[a, b]}=\frac{s(b)-s(a)}{b-a} $$ More recently in Chapter \(4,\) we found that for an object moving along a straight line with velocity function \(v(t),\) the object's "average value of its velocity function on \([a, b]^{\prime \prime}\) is $$ v_{\mathrm{AVG}[a, b]}=\frac{1}{b-a} \int_{a}^{b} v(t) d t $$ Are the "average velocity on the interval \([a, b]^{\prime \prime}\) and the "average value of the velocity function on \([a, b]^{\prime \prime}\) the same thing? Why or why not? Explain.

On a sketch of \(y=e^{x},\) represent the left Riemann sum with \(n=2\) approximating \(\int_{2}^{3} e^{x} d x\). Write out the terms of the sum, but do not evaluate it: Sum \(=\) ___________________ + ___________________ On another sketch, represent the right Riemann sum with \(n=2\) approximating \(\int_{2}^{3} e^{x} d x\). Write out the terms of the sum, but do not evaluate it: Sum \(=\) ___________________ + ___________________ Which sum is an overestimate? Which sum is an underestimate?

When an aircraft attempts to climb as rapidly as possible, its climb rate (in feet per minute) decreases as altitude increases, because the air is less dense at higher altitudes. Given below is a table showing performance data for a certain single engine aircraft, giving its climb rate at various altitudes, where \(c(h)\) denotes the climb rate of the airplane at an altitude \(h\). $$ \begin{array}{lllllllllll} \hline h \text { (feet) } & 0 & 1000 & 2000 & 3000 & 4000 & 5000 & 6000 & 7000 & 8000 & 9000 & 10,000 \\ \hline c(\mathrm{ft} / \mathrm{min}) & 925 & 875 & 830 & 780 & 730 & 685 & 635 & 585 & 535 & 490 & 440 \\ \hline \end{array} $$ Let a new function called \(m(h)\) measure the number of minutes required for a plane at altitude \(h\) to climb the next foot of altitude. a. Determine a similar table of values for \(m(h)\) and explain how it is related to the table above. Be sure to explain the units. b. Give a careful interpretation of a function whose derivative is \(m(h)\). Describe what the input is and what the output is. Also, explain in plain English what the function tells us. c. Determine a definite integral whose value tells us exactly the number of minutes required for the airplane to ascend to 10,000 feet of altitude. Clearly explain why the value of this integral has the required meaning. d. Use the Riemann sum \(M_{5}\) to estimate the value of the integral you found in (c). Include units on your result.

Let \(S\) be the sum given by $$ S=\left((1.4)^{2}+1\right) \cdot 0.4+\left((1.8)^{2}+1\right) \cdot 0.4+\left((2.2)^{2}+1\right) \cdot 0.4+\left((2.6)^{2}+1\right) \cdot 0.4+\left((3.0)^{2}+1\right) \cdot 0.4 $$ a. Assume that \(S\) is a right Riemann sum. For what function \(f\) and what interval \([a, b]\) is \(S\) an approximation of the area under \(f\) and above the \(x\) -axis on \([a, b] ?\) Why? b. How does your answer to (a) change if \(S\) is a left Riemann sum? a middle Riemann sum? c. Suppose that \(S\) really is a right Riemann sum. What is geometric quantity does \(S\) approximate? d. Use sigma notation to write a new sum \(R\) that is the right Riemann sum for the same function, but that uses twice as many subintervals as \(S .\)

A function \(f\) is given piecewise by the formula $$ f(x)=\left\\{\begin{array}{ll} -x^{2}+2 x+1, & \text { if } 0 \leq x<2 \\ -x+3, & \text { if } 2 \leq x<3 \\ x^{2}-8 x+15, & \text { if } 3 \leq x \leq 5 \end{array}\right. $$ a. Determine the exact value of the net signed area enclosed by \(f\) and the \(x\) -axis on the interval [2,5] b. Compute the exact average value of \(f\) on [0,5] . c. Find a formula for a function \(g\) on \(5 \leq x \leq 7\) so that if we extend the above definition of \(f\) so that \(f(x)=g(x)\) if \(5 \leq x \leq 7,\) it follows that \(\int_{0}^{7} f(x) d x=0\)
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