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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.1.71d

Displacement from a velocity graph Consider the velocity function for an object moving along a line (see figure).
(d) Assuming the velocity remains 10 m/s, for t ≥ 5, find the function that gives the displacement between t = 0 and any time t ≥ 5.
Graph showing velocity in m/s over time in seconds, with a constant velocity of 10 m/s from t=5 seconds onward.

Guida verificata passo dopo passo
1
Step 1: Understand the problem. The goal is to find the displacement function for t ≥ 5, given that the velocity remains constant at 10 m/s from t = 5 onward. Displacement is the integral of velocity over time.
Step 2: Recall the formula for displacement. Displacement is given by the definite integral of the velocity function over the interval of interest. For t ≥ 5, the velocity is constant at 10 m/s, so the integral simplifies.
Step 3: Set up the integral. The displacement from t = 0 to any time t ≥ 5 can be expressed as the sum of two parts: (1) the displacement from t = 0 to t = 5, and (2) the displacement from t = 5 to t. For t ≥ 5, the displacement function is: \( s(t) = \int_{0}^{5} v(t) \, dt + \int_{5}^{t} 10 \, dt \).
Step 4: Evaluate the first integral. From the graph, the velocity function \( v(t) \) changes piecewise between t = 0 and t = 5. Break the integral into segments based on the graph: \( \int_{0}^{5} v(t) \, dt = \int_{0}^{1} 10t \, dt + \int_{1}^{3} 20 \, dt + \int_{3}^{5} (-5t + 35) \, dt \).
Step 5: Evaluate the second integral. For t ≥ 5, the velocity is constant at 10 m/s, so \( \int_{5}^{t} 10 \, dt = 10(t - 5) \). Combine the results from Step 4 and Step 5 to express the total displacement function \( s(t) \).

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Velocity and Displacement Relationship

Velocity is the rate of change of displacement with respect to time. To find displacement over a time interval, one can integrate the velocity function. In this case, since the velocity is constant at 10 m/s for t ≥ 5 seconds, the displacement can be calculated as the product of velocity and time.
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Derivatives Applied To Velocity

Integration in Calculus

Integration is a fundamental concept in calculus used to find the area under a curve, which in the context of velocity, represents displacement. The definite integral of the velocity function from the initial time to a later time gives the total displacement during that interval. For a constant velocity, this simplifies to a straightforward multiplication of velocity and time.
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Fundamental Theorem of Calculus Part 1

Piecewise Functions

A piecewise function is defined by different expressions based on the input value. In this scenario, the velocity function changes at t = 5 seconds, making it a piecewise function. Understanding how to evaluate such functions is crucial for determining displacement over varying intervals of time.
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Piecewise Functions
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