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

문제 15d

3장, 문제 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

검증된 단계별 안내

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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주요 개념

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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.