Table of contents

0. Functions7h 52m
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
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
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
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

4. Applications of Derivatives

Motion Analysis

Calculus

4. Applications of Derivatives

Motion Analysis

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3:41 minutes

Problem 3.4.9

Textbook Question

Free-Fall Applications
Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t² on Mars and s = 11.44t² on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m/sec (about 100 km/h) on each planet?

Verified step by step guidance

Start by understanding the relationship between position, velocity, and acceleration in free fall. The position function s(t) gives the distance fallen over time, and the derivative of this function with respect to time, s'(t), gives the velocity.

For Mars, the position function is s = 1.86t². To find the velocity function, take the derivative of s with respect to t: s'(t) = d/dt(1.86t²).

Calculate the derivative: s'(t) = 2 * 1.86 * t = 3.72t. This is the velocity function for Mars.

Set the velocity function equal to the desired velocity, 27.8 m/sec, and solve for t: 3.72t = 27.8. This will give you the time it takes for the rock to reach 27.8 m/sec on Mars.

Repeat the process for Jupiter. The position function is s = 11.44t². Take the derivative to find the velocity function: s'(t) = 2 * 11.44 * t = 22.88t. Set 22.88t = 27.8 and solve for t to find the time it takes on Jupiter.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Free Fall Motion

Free fall motion refers to the movement of an object under the influence of gravitational force alone, without any other forces acting on it. The distance an object falls is proportional to the square of the time it has been falling, as shown by the equations s = 1.86t² for Mars and s = 11.44t² for Jupiter. Understanding this concept is crucial for analyzing the motion of objects in different gravitational fields.

Velocity in Free Fall

Velocity in free fall is the rate of change of position with respect to time, influenced by gravity. For an object starting from rest, its velocity can be determined by differentiating the position function with respect to time. In this problem, the goal is to find the time it takes for a rock to reach a specific velocity, 27.8 m/sec, on Mars and Jupiter, requiring an understanding of how velocity relates to the position function.

Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which a quantity changes. In the context of free fall, differentiating the position function s = 1.86t² or s = 11.44t² with respect to time gives the velocity function. This process allows us to determine the time required for the rock to reach a given velocity by setting the derivative equal to the desired velocity and solving for time.

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Textbook Question

Motion Along a Coordinate Line

Exercises 1–6 give the positions s = f(t) of a body moving on a coordinate line, with s in meters and t in seconds.

b. Find the body’s speed and acceleration at the endpoints of the interval.

s = 25/t² − 5/t, 1 ≤ t ≤ 5

Textbook Question

Motion Along a Coordinate Line

Exercises 1–6 give the positions s = f(t) of a body moving on a coordinate line, with s in meters and t in seconds.

c. When, if ever, during the interval does the body change direction?

s = 25/(t + 5), −4 ≤ t ≤ 0

Textbook Question

Particle motion At time t, the position of a body moving along the s-axis is s = t³ − 6t² + 9t m.

c. Find the total distance traveled by the body from t = 0 to t = 2.

Textbook Question

Particle motion At time t ≥ 0, the velocity of a body moving along the horizontal s-axis is v = t² − 4t + 3.

b. When is the body moving forward? Backward?

Textbook Question

Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 m/sec (about 86 km/h) reaches a height of s = 24t − 0.8t² m in t sec.

e. How long is the rock aloft?

Textbook Question

Understanding Motion from Graphs

Launching a Rocket When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket coasts upward for a while and then begins to fall. A small explosive charge pops out a parachute shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands.

The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following.

a. How fast was the rocket climbing when the engine stopped?

Textbook Question

Understanding Motion from Graphs

The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following.

b. For how many seconds did the engine burn?

Textbook Question

Understanding Motion from Graphs

The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following.

c. When did the rocket reach its highest point? What was its velocity then?

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