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 19c
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) = 2t3 - 21t2 + 60t; 0 ≤ t ≤ 6

1
Step 1: Understand that the velocity of an object is the derivative of its position function with respect to time. Therefore, to find the velocity function v(t), differentiate the position function f(t) = 2t^3 - 21t^2 + 60t with respect to t.
Step 2: Apply the power rule of differentiation to find the derivative of each term in f(t). The power rule states that d/dt [t^n] = n*t^(n-1).
Step 3: Differentiate the first term: d/dt [2t^3] = 3*2*t^(3-1) = 6t^2.
Step 4: Differentiate the second term: d/dt [-21t^2] = 2*(-21)*t^(2-1) = -42t.
Step 5: Differentiate the third term: d/dt [60t] = 60, since the derivative of t with respect to t is 1. Combine these results to get the velocity function v(t) = 6t^2 - 42t + 60.

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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) = 2t³ - 21t² + 60t provides a mathematical representation of the object's position in feet as a function of time in seconds. Understanding this function is crucial for determining the object's velocity and acceleration.
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Velocity
Velocity is the rate of change of position with respect to time, mathematically represented as the first derivative of the position function, v(t) = f'(t). For the given function, calculating the derivative will yield the velocity at any time t, including t = 1. This concept is essential for understanding how fast the object is moving at a specific moment.
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Acceleration
Acceleration is the rate of change of velocity with respect to time, represented as the second derivative of the position function, a(t) = f''(t). By finding the second derivative of the position function, we can determine the object's acceleration at any time, including t = 1. This concept helps in understanding how the object's velocity is changing over time.
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