Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.1.8d

Chapter 6, Problem 6.1.8d

Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.

d. What is the displacement of the object over the interval [0, 8]?

Verified step by step guidance

Understand that displacement over an interval is the net change in position, which can be found by calculating the definite integral of the velocity function over that interval.

Identify the given areas under the velocity curve between the time intervals: from 0 to 1 (area = 20, below the t-axis), from 2 to 4 (area = 14, above the t-axis), from 5 to 6 (area = 10, below the t-axis), and from 6 to 8 (area = 6, above the t-axis).

Note that areas below the t-axis represent negative velocity (movement in the opposite direction), so these areas contribute negatively to the displacement, while areas above the t-axis contribute positively.

Calculate the net displacement by summing the signed areas: subtract the areas below the t-axis from the areas above the t-axis, i.e., displacement = (14 + 6) - (20 + 10).

Express the displacement as the definite integral of velocity from 0 to 8, which corresponds to the sum of these signed areas, representing the net change in position over the interval [0, 8].

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Key Concepts

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

Velocity and Displacement

Velocity is the rate of change of position with respect to time and can be positive or negative depending on direction. Displacement over a time interval is the net change in position, calculated as the integral of velocity over that interval. It accounts for direction, so areas above and below the time-axis affect displacement differently.

Definite Integral and Area Under the Curve

The definite integral of a velocity function over a time interval represents the net area between the velocity curve and the time-axis. Positive areas correspond to motion in the positive direction, while negative areas correspond to motion in the opposite direction. Summing these signed areas gives the displacement.

Interpreting Signed Areas for Displacement

When calculating displacement from a velocity graph, areas above the t-axis are positive contributions, and areas below are negative. To find total displacement, add positive areas and subtract negative areas. This contrasts with total distance, which sums all areas as positive values.

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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.

e. How far has the racer traveled when it reaches a speed of 178 ft/s?

Textbook Question

Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.

e. Describe the position of the object relative to its initial position after 8 seconds.

Textbook Question

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.

d. A piecewise function for s(t)

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.

d. How long does it take the racer to travel 300 ft?

Textbook Question

Displacement and distance from velocity Consider the graph shown in the figure, which gives the velocity of an object moving along a line. Assume time is measured in hours and distance is measured in miles. The areas of three regions bounded by the velocity curve and the t-axis are also given.

d. What is the displacement of the object over the interval [0,5]?

Textbook Question

Determine whether the following statements are true and give an explanation or counterexample.

d. Let f(x)=12x^2.. The area of the surface generated when the graph of f on [−4, 4] is revolved about the y-axis is twice the area of the surface generated when the graph of f on [0, 4] is revolved about the y-axis.