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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 6.1.30
Chapter 6, Problem 6.1.30

29–36. Position and velocity from acceleration Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. Use the Fundamental Theorem of Calculus (Theorems 6.1 and 6.2).


a(t) = −32; v(0)=50; s(0)=0

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Identify the given acceleration function: \(a(t) = -32\), the initial velocity \(v(0) = 50\), and the initial position \(s(0) = 0\).
Recall that velocity is the integral of acceleration with respect to time: \(v(t) = \int a(t) \, dt + C_1\).
Integrate the acceleration function: \(v(t) = \int -32 \, dt = -32t + C_1\).
Use the initial velocity condition \(v(0) = 50\) to solve for the constant \(C_1\): substitute \(t=0\) into \(v(t)\) and set equal to 50.
Next, find the position function by integrating the velocity function: \(s(t) = \int v(t) \, dt + C_2\). Then use the initial position \(s(0) = 0\) to solve for \(C_2\).

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Key Concepts

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

Acceleration, Velocity, and Position Relationship

Acceleration is the rate of change of velocity with respect to time, and velocity is the rate of change of position. Given acceleration, integrating it once with respect to time yields velocity, and integrating velocity gives position. Initial conditions help determine the constants of integration.
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Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, stating that integration can be reversed by differentiation. It allows us to find a function from its derivative by integrating, which is essential for finding velocity and position from acceleration.
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Fundamental Theorem of Calculus Part 1

Initial Conditions and Constants of Integration

When integrating acceleration to find velocity and position, constants of integration appear. Initial velocity and initial position values are used to solve for these constants, ensuring the solution matches the physical scenario described.
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Related Practice
Textbook Question

35–38. Shell and washer methods Let R be the region bounded by the following curves. Use both the shell method and the washer method to find the volume of the solid generated when R is revolved about the indicated axis.


y = (x−2)³ −2,x=0, and y=25; about the y-axis

Textbook Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis. 


x = x³ ,y = 1, and x = 0; about the x-axis

Textbook Question

Express the area of the shaded region in Exercise 5 as the sum of two integrals with respect to y. Do not evaluate the integrals.

Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

y = x^3/2 / 3 − x^1/2 on [4, 16]

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Textbook Question

Use the general slicing method to find the volume of the following solids.

The solid whose base is the region bounded by the curves y=x^2 and y=2−x^2, and whose cross sections through the solid perpendicular to the x-axis are squares

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Textbook Question

35–38. Shell and washer methods Let R be the region bounded by the following curves. Use both the shell method and the washer method to find the volume of the solid generated when R is revolved about the indicated axis.


y = 8,y = 2x+2,x = 0, and x=2; about the y-axis

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