Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
4. Applications of Derivatives
Motion Analysis
Problem 15c
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

1
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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Derivatives 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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