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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
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.1.37c
Chapter 6, Problem 6.1.37c

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?

Verified step by step guidance
1
Identify the given acceleration function: \(a(t) = 88\) ft/s², which is constant.
Since acceleration is the derivative of velocity, integrate \(a(t)\) with respect to \(t\) to find the velocity function \(v(t)\): \[v(t) = \int a(t) \, dt = \int 88 \, dt = 88t + C_1\] Use the initial condition \(v(0) = 0\) to solve for \(C_1\).
Next, velocity is the derivative of position, so integrate \(v(t)\) to find the position function \(s(t)\): \[s(t) = \int v(t) \, dt = \int 88t \, dt = 44t^2 + C_2\] Use the initial condition \(s(0) = 0\) to solve for \(C_2\).
Convert the distance to consistent units: Since \(1\) mile = \(5280\) feet, \(\frac{1}{4}\) mile = \(\frac{1}{4} \times 5280 = 1320\) feet.
Set the position function equal to \(1320\) feet and solve for \(t\): \[44t^2 = 1320\] Solve this equation to find the time \(t\) it takes to travel \(\frac{1}{4}\) mile.

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Key Concepts

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

Acceleration and Its Relationship to Velocity and Displacement

Acceleration is the rate of change of velocity with respect to time. Given a constant acceleration, velocity can be found by integrating acceleration over time, and displacement can be found by integrating velocity. This relationship allows us to determine how velocity and position evolve from acceleration.
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Using The Acceleration Function Example 1

Integration of Constant Acceleration

When acceleration is constant, velocity increases linearly over time, and displacement follows a quadratic function of time. Specifically, velocity is a(t) multiplied by time plus initial velocity, and displacement is (1/2) times acceleration times time squared plus initial displacement.
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Using The Acceleration Function

Unit Conversion and Distance Measurement

To solve for time when given a distance in miles, it is essential to convert miles to feet to match the units of acceleration (ft/s²). Since 1 mile equals 5280 feet, converting 1/4 mile to feet ensures consistent units for accurate calculation.
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A nonlinear spring Hooke’s law is applicable to idealized (linear) springs that are not stretched or compressed too far from their equilibrium positions. Consider a nonlinear spring whose restoring force is given by F(x) = 16x−0.1x³, for |x|≤7. 

c. How much work is done in compressing the spring from its equilibrium position (x=0) to x=−2?

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Textbook Question

6–8. Let R be the region bounded by the curves y = 2−√x,y=2, and x=4 in the first quadrant.

Suppose the shell method is used to determine the volume of the solid generated by revolving R about the line x=4.

c. Write an integral for the volume of the solid using the shell method.

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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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Textbook Question

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.


c. Find the distance traveled over the given interval.


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Textbook Question

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


Arc length may be negative if f(x) < 0 on part of the interval in question.

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