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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.1.9c

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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c. The position at t=5

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Understand that the position function \(s(t)\) is the integral of the velocity function \(v(t)\) over time, starting from the initial position \(s(0) = 0\).
Identify the intervals on the graph where the velocity function changes. From the graph, note the velocity values and the time intervals: from \(t=0\) to \(t=2\), velocity decreases linearly from 2 to -2; from \(t=2\) to \(t=3\), velocity increases linearly from -2 to 3; from \(t=3\) to \(t=5\), velocity decreases linearly from 3 to 0.
Calculate the area under the velocity curve for each interval, as the integral of velocity over time gives the change in position. Since the graph consists of straight lines, these areas can be found by calculating the areas of geometric shapes (triangles and trapezoids) formed between the velocity curve and the time axis.
Sum the areas from each interval, taking care to consider the sign of the velocity (areas below the time axis represent negative displacement). This sum will give the net change in position from \(t=0\) to \(t=5\).
Add the net change in position to the initial position \(s(0) = 0\) to find the position at \(t=5\), \(s(5)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity and Position Relationship

Velocity is the rate of change of position with respect to time. To find the position at a given time, you integrate the velocity function over the time interval, starting from the initial position. This means the position is the accumulated area under the velocity-time graph.
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Derivatives Applied To Velocity

Definite Integral as Area Under the Curve

The definite integral of a velocity function from time a to b represents the net displacement during that interval. Graphically, this corresponds to the area between the velocity curve and the time axis, where areas above the axis are positive and below are negative.
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Definition of the Definite Integral

Piecewise Linear Functions and Area Calculation

When the velocity graph is piecewise linear, the area under the curve can be found by calculating areas of simple geometric shapes like triangles and rectangles. Summing these areas with appropriate signs gives the net displacement, which is used to find the position.
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Piecewise Functions
Related Practice
Textbook Question

Region R is revolved about the line y=1 to form a solid of revolution.


c. Write an integral for the volume of the solid.

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Textbook Question

Let R be the region bounded by the curve y=√cos x and the x-axis on [0, π/2]. A solid of revolution is obtained by revolving R about the x-axis (see figures). 


c. Write an integral for the volume of the solid.

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Textbook Question

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

c. The work required to lift a 10-kg object vertically 10 m is the same as the work required to lift a 20-kg object vertically 5 m.

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c. What is the cardiac output over a period of 1 min? (Use calculus; then check your answer with algebra.)

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Textbook Question

Depletion of natural resources Suppose r(t) = r0e^−kt, with k>0, is the rate at which a nation extracts oil, where r0=10⁷ barrels/yr is the current rate of extraction. Suppose also that the estimate of the total oil reserve is 2×10⁹ barrels. 


c. Find the minimum decay constant k for which the total oil reserves will last forever.

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Textbook Question

Acceleration A drag racer accelerates at a(t)=88 ft/s². Assume v(0)=0, s(0)=0, and t is measured in seconds.


c. At this rate, how long will it take the racer to travel 1/4 mi?

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