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 Integrals3h 25m
4. Applications of Derivatives
Related Rates
Problem 2.R.2
Textbook Question
The height above the ground of a stone thrown upwards is given by s(t), where t is measured in seconds. After 1 second, the height of the stone is 48 feet above the ground, and after 1.5 seconds, the height of the stone is 60 feet above the ground. Evaluate s(1) and s(1.5), and then find the average velocity of the stone over the time interval [1, 1.5].
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1
Identify the given values: s(1) = 48 feet and s(1.5) = 60 feet, which represent the height of the stone at t = 1 second and t = 1.5 seconds respectively.
Recall the formula for average velocity over an interval [a, b], which is given by the formula: \( v_{avg} = \frac{s(b) - s(a)}{b - a} \).
Substitute the known values into the average velocity formula: let a = 1 and b = 1.5, so you will have: \( v_{avg} = \frac{s(1.5) - s(1)}{1.5 - 1} \).
Plug in the heights: \( v_{avg} = \frac{60 - 48}{1.5 - 1} \).
Simplify the expression to find the average velocity: calculate the difference in heights and the difference in time.
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