Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s \u003e 0 corresponding to positions right of the origin.Determine the acceleration of the object when its velocity is zero.f(t) = t2 - 4t; 0 ≤ t ≤ 5

Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 15d

Chapter 3, Problem 15d

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.
Determine the acceleration of the object when its velocity is zero.
f(t) = t² - 4t; 0 ≤ t ≤ 5

Verified step by step guidance

Step 1: Find the velocity function by differentiating the position function f(t) with respect to time t. The velocity function v(t) is the first derivative of f(t), so v(t) = f'(t).

Step 2: Differentiate f(t) = t^2 - 4t to find v(t). This gives v(t) = 2t - 4.

Step 3: Set the velocity function v(t) equal to zero to find the time(s) when the velocity is zero. Solve the equation 2t - 4 = 0 for t.

Step 4: Solve the equation 2t - 4 = 0 to find the value of t when the velocity is zero. This will give you the specific time at which the velocity is zero.

Step 5: Find the acceleration function by differentiating the velocity function v(t) with respect to time t. The acceleration function a(t) is the second derivative of f(t), so a(t) = v'(t). Differentiate v(t) = 2t - 4 to find a(t).

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Key Concepts

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

Position, Velocity, and Acceleration

In calculus, the position of an object is described by a function s = f(t). The velocity is the first derivative of the position function, f'(t), indicating how fast the position changes over time. Acceleration is the second derivative, f''(t), representing the rate of change of velocity. Understanding these relationships is crucial for analyzing motion.

Finding Critical Points

To determine when the velocity is zero, we need to find the critical points of the velocity function. This involves setting the first derivative, f'(t), equal to zero and solving for t. Critical points indicate where the object may change direction or stop, which is essential for analyzing the object's motion.

Evaluating Derivatives

Calculating the first and second derivatives of the position function is necessary to find the velocity and acceleration. For the given function f(t) = t² - 4t, the first derivative f'(t) gives the velocity, while the second derivative f''(t) provides the acceleration. Evaluating these derivatives at the critical points allows us to determine the acceleration when the velocity is zero.

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Related Practice

Textbook Question

Use definition (1) (p. 133) to find the slope of the line tangent to the graph of f at P.

f(x) = x² - 5; P(3,4)

Textbook Question

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.

Determine the velocity and acceleration of the object at t = 1.

f(t) = t² − 4t; 0 ≤ t ≤ 5

Textbook Question

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.

y = sin⁵x

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Textbook Question

On what intervals is the speed increasing?

f(t) = t² - 4t; 0 ≤ t ≤ 5

Textbook Question

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.

y = (3x+7)¹⁰

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Textbook Question

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.

y = (5x²+11x)^4/3