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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 6.1.37c

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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Related Practice
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.

Textbook Question

{Use of Tech} Oscillating motion A mass hanging from a spring is set in motion, and its ensuing velocity is given by v(t) = 2π cos πt, for t≥0. Assume the positive direction is upward and s(0)=0. 


c. At what times does the mass reach its low point the first three times? 

Textbook Question

Blood flow A typical human heart pumps 70 mL of blood (the stroke volume) with each beat. Assuming a heart rate of 60 beats/min (1 beat/s), a reasonable model for the outflow rate of the heart is V′(t)=70(1+sin 2πt), where V(t) is the amount of blood (in milliliters) pumped over the interval [0,t],V(0)=0 and t is measured in seconds.


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

Textbook Question

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


Region R is revolved about the y-axis to form a solid of revolution whose cross sections are washers.


c. What is the area A(y) of a cross section of the solid at a point y in [1, 3]?

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.


v(t) = 3t²−6t on [0, 3]

Textbook Question

Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.


c. Find the distance traveled by the object on the interval 0≤t≤6.