Problem 98 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

It takes about 1.24 seconds for them to fall 4 ft on the moon.

Step 1: Understand the Problem Statement

We are given a differential equation to model the fall of an object on the moon. Our goal is to find the time it takes for the hammer and feather to reach the ground when they start from a height of 4 ft. The acceleration due to gravity on the moon is given as \(-5.2 \ \text{ft/sec}^2\).

Step 2: Set Up the Differential Equation

The differential equation provided is \(\frac{d^2 s}{d t^2} = -5.2\). This describes the acceleration, which is the second derivative of the position function \(s(t)\). We need to integrate this to find \(s(t)\), the position function.

Step 3: Integrate the Differential Equation

First, integrate \(\frac{d^2 s}{d t^2} = -5.2\) to find the velocity function: \(\frac{d s}{d t} = \int -5.2 \, dt = -5.2t + C_1\). We use the initial condition \(\frac{d s}{d t} = 0\) when \(t = 0\) to find \(C_1 = 0\). So, the velocity function is \(\frac{d s}{d t} = -5.2t\).

Step 4: Integrate the Velocity Function

Now, integrate the velocity function \(\frac{d s}{d t} = -5.2t\) to find the position function: \(s = \int -5.2t \, dt = -2.6t^2 + C_2\). Use the initial condition \(s = 4\) when \(t = 0\) to find \(C_2 = 4\). Thus, the position function is \(s(t) = -2.6t^2 + 4\).

Step 5: Solve for when the Object Hits the Ground

The object hits the ground when \(s(t) = 0\). Set the position equation to zero and solve for \(t\):\(0 = -2.6t^2 + 4\). Rearranging gives \(2.6t^2 = 4\), so \(t^2 = \frac{4}{2.6}\), which simplifies to \(t^2 = \frac{20}{13}\). Taking the square root yields \(t \approx \sqrt{\frac{20}{13}}\approx 1.24\).

Step 6: Answer the Question

Now that we have calculated \(t\), we can conclude that it takes approximately 1.24 seconds for the hammer and feather to fall 4 ft on the moon.

Key Concepts

Constant Acceleration

When examining motion, especially in physics, **constant acceleration** plays a crucial role. Constant acceleration means that the rate of change of velocity of an object does not change over time. In simpler terms, the object's speed is changing at a steady rate. In the context of our exercise, the hammer and feather fall towards the moon's surface with a constant acceleration of \(-5.2 \ \text{ft/sec}^2\).This steady acceleration is pivotal because it simplifies calculations. Rather than needing to account for variations, the motion can be described using straightforward mathematical formulas.

• The differential equation \(\frac{d^2 s}{d t^2} = -5.2\) represents this constant acceleration.
• By recognizing that acceleration is the second derivative of the position function \(s(t)\), we can use integration to unravel the motion of the hammer and feather.

In environments without atmospheric drag, like the moon, constant acceleration ensures that all objects, regardless of mass, fall at the same rate. That's why both the hammer and the feather hit the ground simultaneously.

Initial Value Problem

An **initial value problem** (IVP) is a type of problem where we start with a differential equation and a set of initial conditions. The initial conditions provide specific values for the unknown function and its derivatives at the start, or initial time.For our problem:

• The given differential equation is \(\frac{d^2 s}{d t^2} = -5.2\).
• The initial conditions specify that \(\frac{d s}{d t} = 0\) (initial velocity) and \(s = 4\) (initial position) when \(t = 0\).

These conditions simplify the process of finding a particular solution. By integrating the differential equation and applying the initial conditions, we can derive the velocity and position functions.For instance, integrating gives us the velocity equation, which is further integrated to provide the position equation. Using the initial values helps us find specific constants in these equations, ensuring they accurately represent the actual physical motion being analyzed.

Lunar Gravity

**Lunar gravity** refers to the gravitational field experienced on the surface of the moon. It is significantly weaker than Earth's gravity, approximately \(1/6\) of Earth's gravitational force. This exercise highlights lunar gravity, with a specified acceleration of \(-5.2 \ \text{ft/sec}^2\).This reduced gravitational force results in slower falling speeds for objects on the moon compared to Earth. Without atmospheric interference, all objects fall at this uniform rate. The situation described in the exercise demonstrates how lunar gravity affects free-fall motion.Understanding lunar gravity helps us comprehend:

• Why objects fall slower on the moon, as seen in the astronaut's experiment.
• The need to adjust calculations of motion dynamics due to these differing gravitational conditions.

Being aware of such variations is crucial for astronauts operating on the moon and is fundamental in planning lunar missions and activities.