Problem 11 - Thomas Calculus

Question

Finding \(g\) on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of \(15 \mathrm{m} / \mathrm{sec} .\) Because the acceleration of gravity at the planet's surface was \(g_{s} \mathrm{m} / \mathrm{sec}^{2}\) , the explorers expected the ball bearing to reach a height of \(s=15 t-(1 / 2) g_{s} t^{2}\) meters \(t\) sec later. The ball bearing reached its maximum height 20 sec after. The ball bearing was the value of \(g_{s} ?\)

Step-by-Step Solution

Answer

The value of \( g_s \) is 0.75.

Step 1: Recognize the problem statement

The problem involves launching a ball bearing vertically on a planet with unknown gravity, described by the equation of motion: \( s = 15t - \frac{1}{2} g_s t^2 \). We know it reaches its maximum height after 20 seconds.

Step 2: Identify the maximum height condition

The ball reaches its maximum height when its velocity is zero. The velocity function is derived by differentiating the height function: \( v(t) = \frac{ds}{dt} = 15 - g_s t \). At the maximum height, \( v(t) = 0 \).

Step 3: Set the velocity to zero

We have: \( 15 - g_s t = 0 \). Given that this occurs at \( t = 20 \) sec, substitute into the equation to find: \( 15 - g_s \cdot 20 = 0 \).

Step 4: Solve for \(g_s\)

Rearrange the equation \( 15 - g_s \cdot 20 = 0 \) to find \( g_s \). This gives: \( g_s = \frac{15}{20} \) resulting in \( g_s = 0.75 \).

Step 5: Verify the result

Check if \( g_s = 0.75 \) satisfies the conditions set in the problem. Substitute back into the original height equation and verify that it results in a maximum height occurring at \( t = 20 \) seconds.

Key Concepts

Equations of Motion

Equations of motion are mathematical expressions that describe how objects move under the influence of forces. In the problem above, we encounter an equation of motion for a ball bearing being launched vertically upward on a planet without an atmosphere. This is given by:

• Height as a function of time: \[ s = 15t - \frac{1}{2} g_s t^2 \]

This equation tells us that the height \(s\) is a function of time \(t\), the initial velocity (15 m/s), and the gravitational acceleration \(g_s\) on the planet. Initially, the ball bearing gains height, but over time, gravity slows it down until it stops momentarily at its maximum height. Understanding this equation's components is essential as they showcase the relationships between distance, time, and acceleration in vertical motion. By analyzing these equations, explorers can determine unknown values like \(g_s\), helping them characterize planetary gravitational fields.

Differentiation in Calculus

Differentiation is a fundamental concept in calculus that allows us to determine the rate of change of a function. In this context, we use differentiation to derive the velocity function of the ball bearing from its height equation. The original height equation is represented as:

• \( s = 15t - \frac{1}{2} g_s t^2 \)

By differentiating this equation with respect to time \(t\), we obtain the velocity function:

• Velocity as a function of time: \( v(t) = \frac{ds}{dt} = 15 - g_s t \)

This velocity function tells us how the speed of the ball bearing changes over time as it travels upward against gravity. Differentiation here is crucial because it provides insights into when the object reaches its maximum height (velocity equals zero). At this point, explorers use calculus to determine the gravitational force \(g_s\) and to predict the ball bearing's behavior under these conditions.

Velocity and Acceleration

Velocity and acceleration are key concepts that describe motion. Velocity illustrates the speed and direction of motion, while acceleration refers to changes in that velocity. When we launch the ball bearing upwards, its initial velocity is given as 15 m/s.

• Initial velocity: 15 m/s.

As it ascends, the ball bearing slows down due to the acceleration due to gravity, \(g_s\). The velocity equation

• \( v(t) = 15 - g_s t \)

helps us understand this relationship. As the ball bearing climbs higher, gravity decelerates it until it stops (where \(v(t) = 0\)), defining its maximum height. Here, acceleration shows us how the gravitational force works against initial velocity. This understanding is critical to predicting and interpreting object movement in various gravitational fields.