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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 4.7.101c
Chapter 4, Problem 4.7.101c

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.
iii. Now find the body’s displacement from t = 1 to t = 3 given that s = s₀ when t = 0.

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1
Identify the given velocity function: \(v(t) = \frac{ds}{dt} = 9.8t - 3\).
Recall that displacement over the time interval \([t_1, t_2]\) is found by integrating the velocity function over that interval: \(\Delta s = \int_{t_1}^{t_2} v(t) \, dt\).
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 of the velocity function: \(\int (9.8t - 3) \, dt = 4.9t^2 - 3t + C\), where \(C\) is the constant of integration.
Evaluate the definite integral by computing \(\left[4.9t^2 - 3t\right]_{1}^{3} = (4.9 \times 3^2 - 3 \times 3) - (4.9 \times 1^2 - 3 \times 1)\) to find the displacement.

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Key Concepts

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

Velocity and Displacement Relationship

Velocity is the rate of change of displacement with respect to time. Displacement over a time interval can be found by integrating the velocity function over that interval, which accumulates the total change in position.
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Definite Integration

Definite integration calculates the net area under a curve between two limits. In this context, integrating the velocity function from t = 1 to t = 3 gives the total displacement during that time period.
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Initial Conditions and Antiderivatives

An antiderivative of velocity gives the displacement function plus a constant. Using the initial condition s = s₀ at t = 0 allows determination of this constant, enabling calculation of displacement at any time.
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