Skip to main content
Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.1.13b
Capitolo 6, Problema 6.1.13b

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.


b. Find the displacement over the given interval. 


v(t) = 3t²−6t on [0, 3]

Guida verificata passo dopo passo
1
Identify the given velocity function: \(v(t) = 3t^{2} - 6t\) and the time interval \([0, 3]\) seconds.
Recall that displacement over a time interval \([a, b]\) is found by integrating the velocity function over that interval: \(\text{Displacement} = \int_{a}^{b} v(t) \, dt\).
Set up the definite integral for displacement: \(\int_{0}^{3} (3t^{2} - 6t) \, dt\).
Integrate the function term-by-term: the integral of \$3t^{2}$ is $t^{3}$, and the integral of $-6t$ is $-3t^{2}$, so the antiderivative is $t^{3} - 3t^{2}$.
Evaluate the antiderivative at the bounds and subtract: calculate \(\left[ t^{3} - 3t^{2} \right]_{0}^{3} = (3^{3} - 3 \times 3^{2}) - (0 - 0)\) to find the displacement.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
2m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Velocity and Displacement Relationship

Velocity represents the rate of change of position with respect to time. Displacement over a time interval is the net change in position, which can be found by integrating the velocity function over that interval. This integral sums the instantaneous velocities to give the overall change in position.
Video consigliato:
Percorso guidato
06:29
Derivatives Applied To Velocity

Definite Integral as Net Area

The definite integral of a velocity function over a time interval calculates the net area between the velocity curve and the time axis. Positive areas correspond to movement in one direction, while negative areas represent movement in the opposite direction. The sum of these areas gives the total displacement.
Video consigliato:
Percorso guidato
05:43
Definition of the Definite Integral

Evaluating Definite Integrals

To find displacement, compute the definite integral of v(t) from the start to end time. This involves finding the antiderivative of v(t), then applying the Fundamental Theorem of Calculus by subtracting the antiderivative evaluated at the lower limit from that at the upper limit.
Video consigliato:
Percorso guidato
05:43
Definition of the Definite Integral
Pratica correlata
Domanda del libro di testo

Equal integrals Without evaluating integrals, explain the following equalities. (Hint: Draw pictures.)


b. ∫²₀(25−(x²+1)²) dx = 2∫₁⁵ y√y−1 dy

110
views
Domanda del libro di testo

Different axes of revolution Suppose R is the region bounded by y=f(x) and y=g(x) on the interval [a, b], where f(x)≥g(x).


b. How is this formula changed if x0>b?

38
views
Domanda del libro di testo

Calculating work for different springs Calculate the work required to stretch the following springs 0.4 m from their equilibrium positions. Assume Hooke’s law is obeyed.

b. A spring that requires 2 J of work to be stretched 0.1 m from its equilibrium position

82
views
Domanda del libro di testo

Deceleration A car slows down with an acceleration of a(t) = −15 ft/s². Assume v(0)=60 ft/s,s(0)=0, and t is measured in seconds.


b. How far does the car travel in the time it takes to come to rest?

31
views
Domanda del libro di testo

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


b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal.

34
views
Domanda del libro di testo

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


b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

42
views