Skip to main content
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 6.1.8d
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.
img
d. What is the displacement of the object over the interval [0, 8]? 

Verified step by step guidance
1
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].

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

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.
Recommended video:
10:17
Using The Velocity Function

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.
Recommended video:
05:43
Definition of the Definite Integral

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.
Recommended video:
05:06
Finding Area When Bounds Are Not Given
Related Practice
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 x-axis to form a solid of revolution whose cross sections are washers.


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

126
views
Textbook Question

Compressing and stretching a spring Suppose a force of 30 N is required to stretch and hold a spring 0.2 m from its equilibrium position.

d. How much additional work is required to stretch the spring 0.2m if it has already been stretched 0.2m from its equilibrium position?

59
views
Textbook Question

Piecewise velocity The velocity of a (fast) automobile on a straight highway is given by the function

v(t)={3t if 0≤t<2060 if 20≤t<45240−4t if t≥45v(t)= \(\begin{cases}\)3 t & \(\text\) { if } 0 \(\leq\) t<20 \\ 60 & \(\text\) { if } 20 \(\leq\) t<45 \\ 240-4 t & \(\text\) { if } t \(\geq\) 45\(\end{cases}\)

where is measured in seconds and v has units of m/s. 

d. What is the position of the automobile when t=75?

48
views
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?

27
views
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]?

117
views
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.

56
views