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 17d
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) = 2t2 - 9t + 12; 0 ≤ t ≤ 3

1
Step 1: Find the velocity function by differentiating the position function s = 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) = 2t^2 - 9t + 12 to find v(t). Use the power rule for differentiation: if f(t) = at^n, then f'(t) = nat^(n-1).
Step 3: Set the velocity function v(t) equal to zero to find the time(s) when the velocity is zero. Solve the equation v(t) = 0 for t.
Step 4: 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) = f''(t).
Step 5: Evaluate the acceleration function a(t) at the time(s) found in Step 3 to determine the acceleration of the object when its velocity is zero.

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mPlay a video:
Was this helpful?
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 function s = f(t) describes the location of an object over time. The velocity is the first derivative of the position function, indicating how fast the position changes with respect to time. Acceleration, the second derivative of the position function, measures how the velocity changes over time. Understanding these relationships is crucial for analyzing motion.
Recommended video:
Derivatives Applied To Acceleration
Finding Critical Points
To determine when the velocity is zero, we need to find the critical points of the velocity function, which is derived from the position function. This involves setting the first derivative (velocity) equal to zero and solving for t. Critical points indicate where the object may change direction or stop, which is essential for analyzing motion.
Recommended video:
Critical Points
Second Derivative Test
The second derivative test helps determine the nature of critical points found in the first derivative. By evaluating the second derivative at these points, we can ascertain whether the object is accelerating or decelerating. Specifically, if the second derivative is positive, the object is accelerating; if negative, it is decelerating. This is vital for understanding the object's motion when its velocity is zero.
Recommended video:
The Second Derivative Test: Finding Local Extrema
Watch next
Master Derivatives Applied To Velocity with a bite sized video explanation from Nick
Start learning