/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A dropped stone on Mars A stone ... [FREE SOLUTION] | 91Ó°ÊÓ

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A dropped stone on Mars A stone is dropped off the edge of a 54 -ft cliff on Mars, where the acceleration due to gravity is about \(12 \mathrm{ft} / \mathrm{s}^{2} .\) The height (in feet) of the stone above the ground \(t\) seconds after it is dropped is \(s(t)=-6 t^{2}+54 .\) Find the velocity of the stone and its speed when it hits the ground.

Short Answer

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Answer: v(t) = -12t 2. After how many seconds does the stone hit the ground? Answer: 3 seconds 3. What is the velocity of the stone when it hits the ground? Answer: -36 ft/s 4. What is the speed of the stone when it hits the ground? Answer: 36 ft/s

Step by step solution

01

Find the velocity function

To find the velocity function, we will take the derivative of the height function \(s(t)\) with respect to time. This will give us: \(v(t) = \frac{\mathrm{d}s(t)}{\mathrm{d}t} = \frac{\mathrm{d}(-6t^2 + 54)}{\mathrm{d}t}\) Now, calculate the derivative: \(v(t) = -12t\) The velocity function is \(v(t) = -12t\). This function tells us the velocity of the stone at any given time.\(t\).
02

Find when the stone hits the ground

When the stone hits the ground, its height will be 0. We can find when this happens by solving the height function \(s(t)\) for \(t\) when \(s(t) = 0\): \(0 = -6t^2 + 54\) Now, solve for \(t\): \(t^2 = 9\) \(t = \pm3\) Since the time must be positive, the stone hits the ground at \(t = 3\) seconds.
03

Find the velocity and speed when the stone hits the ground

Using the time we found in Step 2, we can find the velocity and speed when the stone hits the ground. First, find the velocity: \(v(3) = -12(3) = -36 \, \mathrm{ft} / \mathrm{s}\) The velocity of the stone when it hits the ground is \(-36 \, \mathrm{ft}/\mathrm{s}\). The negative sign indicates that the stone is moving downward. Now, find the speed: Since speed is the absolute value of the velocity, the speed of the stone when it hits the ground is: \(|v(3)| = |-36 \, \mathrm{ft}/\mathrm{s}| = 36 \, \mathrm{ft}/\mathrm{s}\) The stone's speed when it hits the ground is 36 ft/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus in Physics
Physics often deals with changing quantities - whether it's the position of a planet or the speed of a falling object. Calculus is the mathematical tool used to analyze and calculate these changes. In physics, calculus enables us to determine how these quantities evolve over time or space.

When dealing with motion, such as that of a stone falling on Mars, we use calculus to find how the position and velocity change. The height function, often denoted as \( s(t) \), describes the position of an object at any given time. Calculus allows us to take the derivative of this function to get the velocity function \( v(t) \), which indicates how fast the object is moving and in which direction at any point in time. Integrating the velocity function, on the other hand, gives us the displacement of the object over a period. This integration and differentiation are fundamental concepts of calculus when applied to physics.
Velocity Function
The velocity function is a core concept in kinematics, which is a branch of physics that describes the motion of objects without considering the forces causing the motion. The velocity function \( v(t) \) provides us with the velocity of an object at any instant in time.

A velocity function is derived from the position function, or height function in the case of vertical motion, by taking its first derivative with respect to time. This derivative gives us the rate of change of position, which is precisely what velocity measures. In the Mars gravity acceleration problem, the velocity function was found by differentiating the height function \( s(t) = -6t^2 + 54 \) to get \( v(t) = -12t \). This simplified expression makes it easy to plug in any value of time to find the stone's velocity at that exact moment. Understanding the velocity function is crucial for analyzing motion and predicting future positions of moving objects.
Derivative of Height Function
In the context of a free-falling object, the height function describes how the height of the object changes over time. Taking the derivative of the height function gives us the velocity function, a practical example of how derivatives work in calculus to reveal rates of change.

For an object in free fall on Mars, if we denote the height function as \( s(t) \), the velocity function is its derivative \( v(t) = \frac{ds(t)}{dt} \). This mathematical process essentially tells us the object's instantaneous rate of free fall. In the textbook exercise, the process of finding the derivative is executed by applying calculus rules to get the result \( v(t) = -12t \), indicating that the stone's velocity increases linearly with time due to a constant acceleration - in this case, Mars' gravity.
Free Fall on Mars
Free fall on Mars, or any celestial body, refers to the motion of an object under the influence of that body's gravitational field alone. The acceleration due to gravity on Mars is less than on Earth, about \(12 \mathrm{ft/s}^2\), and this affects how fast objects fall. Using calculus, we can model and analyze the free-fall motion of objects on Mars.

In the example problem, we analyzed the free fall of a stone off a cliff. By setting up a height function that models Mars's gravity, we were able to compute the time it takes for the stone to hit the ground and its velocity upon impact. The crucial insight here is that free fall on Mars or any planet follows predictable physical laws, and calculus provides us with the tools to quantify these motions precisely. Understanding the free-fall motion on different planets is not only essential for solving textbook physics problems but also for real-world applications in aeronautics and space exploration.

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Most popular questions from this chapter

An equilateral triangle initially has sides of length \(20 \mathrm{ft}\) when each vertex moves toward the midpoint of the opposite side at a rate of \(1.5 \mathrm{ft} / \mathrm{min}\). Assuming the triangle remains equilateral, what is the rate of change of the area of the triangle at the instant the triangle disappears?

Position, velocity, and acceleration 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. a. Graph the position function. b. Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left? c. Determine the velocity and acceleration of the object at \(t=1\) d. Determine the acceleration of the object when its velocity is zero. e. On what intervals is the speed increasing? $$f(t)=2 t^{3}-21 t^{2}+60 t ; 0 \leq t \leq 6$$

A rectangular swimming pool \(10 \mathrm{ft}\) wide by \(20 \mathrm{ft}\) long and of uniform depth is being filled with water. a. If \(t\) is elapsed time, \(h\) is the height of the water, and \(V\) is the volume of the water, find equations relating \(V\) to \(h\) and \(d V / d t\) to \(d h / d t\) B. At what rate is the volume of the water increasing if the water level is rising at \(\frac{1}{4} \mathrm{ft} / \mathrm{min} ?\) c. At what rate is the water level rising if the pool is filled at a rate of \(10 \mathrm{ft}^{3} / \mathrm{min} ?\)

Demand and elasticity The economic advisor of a large tire store proposes the demand function \(D(p)=\frac{1800}{p-40},\) where \(D(p)\) is the number of tires of one brand and size that can be sold in one day at a price \(p\) a. Recalling that the demand must be positive, what is the domain of this function?b. According to the model, how many tires can be sold in a day at a price of \(\$ 60\) per tire? c. Find the elasticity function on the domain of the demand function. d. For what prices is the demand elastic? Inelastic? e. If the price of tires is raised from \(\$ 60\) to \(\$ 62,\) what is the approximate percentage decrease in demand (using the elasticity function)?

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