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)
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 15c
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
Verified step by step guidance1
Step 1: Understand the problem. We are given a position function s = f(t) = t^2 - 4t, which describes the position of an object moving along a line. We need to find the velocity and acceleration of the object at t = 1.
Step 2: Find the velocity function. The velocity of an object is the derivative of the position function with respect to time. So, calculate the derivative of f(t) with respect to t: v(t) = f'(t) = d/dt (t^2 - 4t).
Step 3: Calculate the derivative. Use the power rule to differentiate each term: the derivative of t^2 is 2t, and the derivative of -4t is -4. Therefore, v(t) = 2t - 4.
Step 4: Evaluate the velocity at t = 1. Substitute t = 1 into the velocity function: v(1) = 2(1) - 4.
Step 5: Find the acceleration function. The acceleration is the derivative of the velocity function. Differentiate v(t) = 2t - 4 with respect to t 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 Function
The position function, denoted as s = f(t), describes the location of an object at any given time t. In this case, the function f(t) = t² - 4t represents the position of the object moving along a horizontal line. Understanding this function is crucial for determining how the object's position changes over time.
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Relations and Functions
Velocity
Velocity is the rate of change of position with respect to time, mathematically represented as the derivative of the position function. To find the velocity at a specific time, we compute f'(t), which gives us the instantaneous rate of change of the position at that moment. For this problem, we need to evaluate the derivative at t = 1.
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06:29Derivatives Applied To Velocity
Acceleration
Acceleration is the rate of change of velocity with respect to time, which can be found by taking the derivative of the velocity function. In this context, it is essential to compute the second derivative of the position function, f''(t), to determine how the velocity of the object changes over time. Evaluating this at t = 1 will provide the acceleration at that instant.
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06:15Derivatives Applied To Acceleration
Related Practice
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.
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
7–14. Find the derivative the following ways:
a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.
y = x² - 2ax +a² / x-a, where a is a constant
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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
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 acceleration of the object when its velocity is zero.
f(t) = t2 - 4t; 0 ≤ t ≤ 5