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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 6.1.10b

9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.
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b. The distance traveled between t=0 and t=5

Guida verificata passo dopo passo
1
Identify the velocity function from the graph. From t = 0 to t = 3, the velocity is constant at 3 units per time. From t = 3 to t = 5, the velocity decreases linearly from 3 to 0.
Recall that the distance traveled is the total length of the path, which is the integral of the absolute value of velocity over the time interval. Since velocity is positive here, distance traveled equals the integral of velocity from t = 0 to t = 5.
Break the integral into two parts corresponding to the two segments of the velocity graph: from 0 to 3 and from 3 to 5. So, calculate \( \int_0^3 3 \, dt \) and \( \int_3^5 v(t) \, dt \), where \( v(t) \) is the linear function decreasing from 3 to 0.
For the second part, find the equation of the line for velocity between t = 3 and t = 5. Use the two points (3, 3) and (5, 0) to find the slope and write \( v(t) = m t + b \).
Calculate each integral separately and then add the results to find the total distance traveled between t = 0 and t = 5.

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Velocity and Position Relationship

Velocity is the rate of change of position with respect to time. The position function s(t) can be found by integrating the velocity function v(t). Understanding this relationship allows us to determine how far an object has moved over a time interval.
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Derivatives Applied To Velocity

Distance Traveled vs. Displacement

Distance traveled is the total length of the path taken, regardless of direction, while displacement is the net change in position. When velocity changes sign, distance traveled is found by integrating the absolute value of velocity, ensuring all movement contributes positively.
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Using The Acceleration Function Example 1

Area Under the Velocity Curve

The area under the velocity-time graph between two times represents displacement. For distance traveled, the total area between the curve and the time axis is considered, treating areas below the axis as positive. This geometric interpretation helps calculate distance from velocity graphs.
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Estimating the Area Under a Curve with Right Endpoints & Midpoint
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Deceleration A car slows down with an acceleration of a(t) = −15 ft/s². Assume v(0)=60 ft/s,s(0)=0, and t is measured in seconds.


b. How far does the car travel in the time it takes to come to rest?

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b. What is the SAV ratio of a ball with radius a? 

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Use the region R that is bounded by the graphs of y=1+√x,x=4, and y=1 complete the exercises.


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b. What is the inner radius of a cross section of the solid at a point y in [1, 3]?

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Consider the following curves on the given intervals.  


b. Use a calculator or software to approximate the surface area.


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40–43. Population growth


When records were first kept (t=0), the population of a rural town was 250 people. During the following years, the population grew at a rate of P′(t) = 30(1+√t), where t is measured in years.


b. Find the population P(t) at any time t≥0.

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Domanda del libro di testo

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal.

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