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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.7.101a
Chapter 4, Problem 4.7.101a

Finding displacement from an antiderivative of velocity


a. Suppose that the velocity of a body moving along the s-axis is


ds/dt = v = 9.8t − 3.


i. Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0.

Verified step by step guidance
1
Identify the given velocity function: \(v(t) = \frac{ds}{dt} = 9.8t - 3\).
Recall that displacement over the interval \([t_1, t_2]\) is given by the change in position \(s(t_2) - s(t_1)\), which can be found by integrating the velocity function over that interval.
Set up the definite integral for displacement from \(t=1\) to \(t=3\): \(\Delta s = \int_{1}^{3} (9.8t - 3) \, dt\).
Find the antiderivative (indefinite integral) of the velocity function: \(S(t) = \int (9.8t - 3) \, dt = 4.9t^2 - 3t + C\), where \(C\) is the constant of integration.
Use the initial condition \(s(0) = 5\) to solve for \(C\) by substituting \(t=0\) into \(S(t)\), then evaluate \(S(3) - S(1)\) to find the displacement over the interval \([1,3]\).

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Key Concepts

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

Relationship Between Velocity and Displacement

Velocity is the rate of change of displacement with respect to time, expressed as v = ds/dt. To find displacement over a time interval, we integrate the velocity function over that interval, which gives the net change in position.
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Derivatives Applied To Velocity

Definite Integration to Find Displacement

Definite integration of the velocity function from the initial time to the final time calculates the total displacement. This integral sums the instantaneous velocities over time, accounting for direction and magnitude of movement.
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Definition of the Definite Integral

Initial Conditions and Antiderivatives

An antiderivative of velocity gives the displacement function s(t) up to a constant. Using the initial condition s(0) = 5 allows us to solve for this constant, ensuring the displacement function accurately reflects the body's position at all times.
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Related Practice
Textbook Question

53. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.

a. Start counting time with t=0 at noon and express the distance s between the ships as a function of t.

Textbook Question

Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


k(x) = x³ + 3x² + 3x + 1, −∞ < x ≤ 0

Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

−πsin πx

Textbook Question

Analyzing Functions from Derivatives


Answer the following questions about the functions whose derivatives are given in Exercises 1–14:


a. What are the critical points of f?


f′(x) = x(x − 1)

Textbook Question

Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


f(t) = 12t − t³, −3 ≤ t < ∞

Textbook Question

25. Paper folding A rectangular sheet of 8.5-in.-by-11-in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper.

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a. Show that L^2=2x^3/(2x-8.5).