At , a car approaching a stop sign decelerates from a speed of 50 / according to the acceleration function , where and is measured in hours. How far does the car travel between and ?
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 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
10. Physics Applications of Integrals
Kinematics
Struggling with Calculus?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
A rock is thrown from a height of 2ft with an initial speed of 25ft/s. Acceleration resulting from gravity is −32ft/s2.
Find
A
B
C
D

1
Step 1: Understand the problem. The rock's motion is described by a quadratic equation derived from the kinematic formula for position under constant acceleration. The general formula is s(t) = s₀ + v₀t + (1/2)at², where s₀ is the initial position, v₀ is the initial velocity, and a is the acceleration due to gravity.
Step 2: Identify the given values. From the problem, the initial height s₀ is 2 ft, the initial velocity v₀ is 25 ft/s, and the acceleration due to gravity a is -32 ft/s². Note that the acceleration is negative because gravity acts downward.
Step 3: Substitute the given values into the kinematic formula. Replace s₀ with 2, v₀ with 25, and a with -32 in the equation s(t) = s₀ + v₀t + (1/2)at².
Step 4: Simplify the equation. Perform the calculation for (1/2) * a, which is (1/2) * (-32) = -16. This gives the equation s(t) = 2 + 25t - 16t².
Step 5: Rearrange the terms to write the equation in standard form. The final equation for the position of the rock as a function of time is s(t) = -16t² + 25t + 2.
Watch next
Master Using The Velocity Function with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice
Multiple Choice