Problem 106 - Thomas Calculus

Question

\(\begin{array}{l}{\text { The hammer and the feather } \text { When Apollo } 15 \text { astronaut }} \\ {\text { David Scott dropped a hammer and a feather on the moon to }} \\ {\text { demonstrate that in a vacuum all bodies fall with the same (con- }} \\ {\text { stant) acceleration, he dropped them from about } 4 \text { ft above the }} \\ {\text { ground. The television footage of the event shows the hammer }}\end{array}$$ \begin{array}{l}{\text { and the feather falling more slowly than on Earth, where, in a }} \\ {\text { vacuum, they would have taken only half a second to fall the } 4} \\ {\text { ft. How long did it take the hammer and feather to fall } 4 \text { ft on }} \\ {\text { the moon? To find out, solve the following initial value prob- }} \\ {\text { lem for } s \text { as a function of } t \text { . Then find the value of } t \text { that makes } s} \\\ {\text { equal to } 0 .}\end{array}\) \(\begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} s}{d t^{2}}=-5.2 \mathrm{ft} / \mathrm{sec}^{2}} \\ {\text { Initial conditions: }} & {\frac{d s}{d t}=0 \text { and } s=4 \text { when } t=0}\end{array}\)

Step-by-Step Solution

Answer

It took about 1.24 seconds for the hammer and feather to fall 4 feet on the Moon.

Step 1: Understand the Problem

We are given a second-order differential equation \( \frac{d^{2}s}{dt^{2}} = -5.2 \; \text{ft/sec}^2 \) representing the acceleration of objects on the Moon. Our task is to find the function \( s(t) \), which describes the position of the hammer and feather over time, using the initial conditions: \( \frac{ds}{dt} = 0 \) and \( s = 4 \) at \( t = 0 \). Then, we need to find the time \( t \) when \( s = 0 \) (when the objects have hit the ground).

Step 2: Integrate the Acceleration to Obtain the Velocity

Integrate the acceleration function \( \frac{d^{2}s}{dt^{2}} = -5.2 \) to obtain the velocity function \( \frac{ds}{dt} \). This gives us: \[ \frac{ds}{dt} = -5.2t + C_1 \] where \( C_1 \) is the integration constant. Using the initial condition \( \frac{ds}{dt} = 0 \) when \( t = 0 \), we find \( C_1 = 0 \). Thus, the velocity function is \( \frac{ds}{dt} = -5.2t \).

Step 3: Integrate the Velocity to Obtain the Position

Integrate the velocity function \( \frac{ds}{dt} = -5.2t \) to find the position function \( s(t) \). This gives us: \[ s(t) = -\frac{5.2}{2}t^2 + C_2 \] where \( C_2 \) is another integration constant. Using the initial condition \( s = 4 \) when \( t = 0 \), we find \( C_2 = 4 \). Thus, the position function is \( s(t) = -2.6t^2 + 4 \).

Step 4: Solve for Time When the Hammer and Feather Hit the Ground

Set \( s(t) = 0 \) to find the time \( t \) when the objects reach the ground: \[ -2.6t^2 + 4 = 0 \] Solving for \( t \) gives: \[ 2.6t^2 = 4 \] \[ t^2 = \frac{4}{2.6} \approx 1.538 \] \[ t = \sqrt{1.538} \approx 1.24 \] seconds. Therefore, it takes approximately 1.24 seconds for the hammer and feather to fall 4 feet on the Moon.

Key Concepts

Initial Value Problem

An initial value problem in differential equations is a fundamental concept where we are given a differential equation and initial conditions to find a particular solution. In this specific exercise, we are given a second-order differential equation which represents the acceleration of a hammer and a feather on the Moon. The problem asks us to solve for the position function \( s(t) \), using initial conditions:

• The initial velocity \( \frac{ds}{dt} \) at \( t = 0 \) is 0.
• The initial position \( s \) is 4 feet when \( t = 0 \).

These initial conditions "anchor" our solution, allowing us to find specific integration constants that tailor the general solution to our specific scenario.

Acceleration on the Moon

Acceleration on the Moon is much weaker than on Earth due to the Moon's lower gravity. In this exercise, the acceleration is given as \( -5.2 \; \text{ft/sec}^2 \). This negative sign indicates that the acceleration is directed downward, towards the Moon's surface.

The experiment performed by astronaut David Scott demonstrated that in a vacuum—like the Moon—different objects fall with the same acceleration, regardless of their mass. This is a perfect context to explore constant acceleration motion where gravity is the sole force acting on the objects.

Integration in Calculus

Integration in calculus is the process of finding a function from its derivative. This is critical in solving differential equations, where we often deal with rates of change.

In our exercise, we first integrate the given acceleration \( \frac{d^{2}s}{dt^{2}} = -5.2 \) to find the velocity \( \frac{ds}{dt} \). This gives us:
\[ \frac{ds}{dt} = -5.2t + C_1 \]
The integration constant \( C_1 \) is determined using the initial condition \( \frac{ds}{dt} = 0 \) when \( t = 0 \), resulting in \( C_1 = 0 \).

We then integrate the velocity to find the position function:
\[ s(t) = -\frac{5.2}{2}t^2 + C_2 \]
The constant \( C_2 \) is found using the condition \( s = 4 \) when \( t = 0 \), which gives us \( C_2 = 4 \). This is how we establish a specific function that describes the motion of the hammer and feather.

Constant Acceleration Motion

Constant acceleration motion involves movements where the rate of change of velocity is uniform over time. This simplifies the calculation of changes in velocity and position.

For this scenario, the constant acceleration is \( -5.2 \; \text{ft/sec}^2 \). The position of the objects falling from a height of 4 feet is expressed by
\[ s(t) = -2.6t^2 + 4 \]
Using this equation, we set \( s(t) = 0 \) to determine when the objects reach the ground, solving it by rearranging:
\[ -2.6t^2 + 4 = 0 \]
The solution gives the time \( t \approx 1.24 \) seconds. This simple example shows how constant acceleration influences motion, allowing precise predictions of position over time.