/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A car comes to a stop six second... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 1

A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded: $$ \begin{array}{|l|l|l|l|l|} \hline \text { Time since brakes applied (sec) } & 0 & 2 & 4 & 6 \\ \hline \text { Velocity (ft/s) } & 88 & 45 & 16 & 0 \\ \hline \end{array} $$ Give lower and upper estimates (using all of the available data) for the distance the car traveled after the brakes were applied. lower: upper: (for each, include units) On a sketch of velocity against time, show the lower and upper estimates you found above..

Short Answer

Expert verified
Lower Estimate: 122 ft, Upper Estimate: 298 ft

Step by step solution

01

- Understanding the Problem

We need to estimate the distance a car travels after the brakes are applied using recorded velocities at specific time intervals. We will use both lower and upper estimates to determine this distance. The data points given are: Time (0, 2, 4, 6) seconds and Velocities (88, 45, 16, 0) ft/s.
02

- Calculating the Lower Estimate

The lower estimate is determined using the minimum speed within each time interval. Calculate the distance for each interval using the formula Distance = Velocity 脳 Time Interval.For 0 to 2 seconds: Velocity = 45 ft/s Distance = 45 脳 2 = 90 ftFor 2 to 4 seconds: Velocity = 16 ft/s Distance = 16 脳 2 = 32 ftFor 4 to 6 seconds: Velocity = 0 ft/s Distance = 0 脳 2 = 0 ftTotal Lower Estimate = 90 + 32 + 0 = 122 ft
03

- Calculating the Upper Estimate

The upper estimate is determined using the maximum speed within each time interval. Calculate the distance for each interval similarly.For 0 to 2 seconds: Velocity = 88 ft/s Distance = 88 脳 2 = 176 ftFor 2 to 4 seconds: Velocity = 45 ft/s Distance = 45 脳 2 = 90 ftFor 4 to 6 seconds: Velocity = 16 ft/s Distance = 16 脳 2 = 32 ftTotal Upper Estimate = 176 + 90 + 32 = 298 ft
04

- Sketching the Velocity vs. Time Graph

Plot the given velocities at their corresponding times on a graph. Draw lines between these points to show changes in velocity over time. The graph will help visualize the lower and upper estimates. The lower estimate uses points (2, 45), (4, 16), and (6, 0), while the upper estimate uses points (0, 88), (2, 45), and (4, 16).

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Key Concepts

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

velocity-time graph
Understanding the velocity-time graph helps us visualize how a car's speed changes over a period. On this graph, the x-axis represents time in seconds, and the y-axis represents velocity in feet per second (ft/s). By plotting the given data points鈥�(0, 88), (2, 45), (4, 16), (6, 0)鈥攚e can see how the car decelerates after braking. Drawing lines between these points shows the deceleration pattern and allows us to estimate the distance traveled.
Riemann sums
To estimate the distance the car travels, we use Riemann sums. These are techniques for approximating the total area under the curve on a graph. The area represents the distance traveled. We calculate two types of Riemann sums: lower and upper estimates.
  • Lower Estimate: Uses the minimum velocities in each time interval.
  • Upper Estimate: Uses the maximum velocities in each time interval.
These estimates give us a range for the possible distance traveled.
lower and upper estimates
Lower and upper estimates provide a range for the distance traveled.
  • **Lower Estimate:** Calculated using the minimum velocity in each time segment.
  • **Upper Estimate:** Calculated using the maximum velocity in each time segment.
For the given problem:
  • Lower Estimate: For time intervals (0-2), (2-4), (4-6) seconds, use velocities 45, 16, 0 ft/s respectively. Calculate the distance for each interval as Distance = Velocity 脳 Time Interval. Total lower estimate: 90 ft + 32 ft + 0 ft = 122 ft.
  • Upper Estimate: For the same intervals, use velocities 88, 45, 16 ft/s respectively. Total upper estimate: 176 ft + 90 ft + 32 ft = 298 ft.
distance calculation
Distance calculation using velocity-time data is straightforward when we break it down. For each time interval, multiply the velocity by the duration of the interval.
Summing these products gives the total distance traveled. In this case:
  • 0 to 2 seconds: Lower estimate uses 45 ft/s, Upper estimate uses 88 ft/s
  • 2 to 4 seconds: Lower estimate uses 16 ft/s, Upper estimate uses 45 ft/s
  • 4 to 6 seconds: Lower and Upper estimates both use 0 and 16 ft/s respectively
By following this method, we accurately find how much distance was covered during the braking period.

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Most popular questions from this chapter

Find the average value of \(f(x)=6 x+5\) over [2,6] average value = _____________

The velocity function is \(v(t)=t^{2}-6 t+8\) for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [-2,5] displacement \(=\) ______________

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.

Find the average value of \(f(x)=4 x+4\) over [4,9] average value \(=\) _________________

Let \(f(x)=3-x^{2}\) and \(g(x)=2 x^{2}\) a. On the interval [-1,1] , sketch a labeled graph of \(y=f(x)\) and write a definite integral whose value is the exact area bounded by \(y=f(x)\) on [-1,1] . b. On the interval \([-1,1],\) sketch a labeled graph of \(y=g(x)\) and write a definite integral whose value is the exact area bounded by \(y=g(x)\) on [-1,1] . c. Write an expression involving a difference of definite integrals whose value is the exact area that lies between \(y=f(x)\) and \(y=g(x)\) on [-1,1] . d. Explain why your expression in (c) has the same value as the single integral \(\int_{-1}^{1}[f(x)-\) \(g(x)] d x\) e. Explain why, in general, if \(p(x) \geq q(x)\) for all \(x\) in \([a, b],\) the exact area between \(y=p(x)\) and \(y=q(x)\) is given by $$ \int_{a}^{b}[p(x)-q(x)] d x $$
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