Problem 124 - University Calculus: Early Transcendentals

Question

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 \(\mathrm{ft}\) above the ground. The television footage 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 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: \(\frac{d s}{d t}=0\) and \(s=4\) when \(t=0\)

Step-by-Step Solution

Answer

It took about 1.24 seconds for the hammer and feather to fall on the moon.

Step 1: Understanding the Problem

We need to find how long it took for the hammer and feather to fall on the moon given the differential equation \( \frac{d^2 s}{d t^2} = -5.2 \text{ ft/sec}^2 \) with initial conditions \( \frac{d s}{d t} = 0 \) and \( s = 4 \) at \( t = 0 \). We are solving for the time \( t \) when the object reaches the ground, i.e., \( s = 0 \).

Step 2: Integrate the Acceleration to Find Velocity

Start by integrating the acceleration function \( \frac{d^2 s}{d t^2} = -5.2 \) to find the velocity function \( \frac{d s}{d t} \). The integration yields \( \frac{d s}{d t} = -5.2t + C_1 \). Since \( \frac{d s}{d t} = 0 \) when \( t = 0 \), substitute into the equation to find \( C_1 = 0 \). Therefore, \( \frac{d s}{d t} = -5.2t \).

Step 3: Integrate Velocity to Find Displacement

Now, integrate \( \frac{d s}{d t} = -5.2t \) to find the displacement \( s \). This gives \( s = -2.6t^2 + C_2 \). Using the initial condition \( s = 4 \) when \( t = 0 \), we find \( C_2 = 4 \). Hence, the displacement function is \( s = -2.6t^2 + 4 \).

Step 4: Solve for Time When Displacement is Zero

Set the displacement equation \( s = -2.6t^2 + 4 \) equal to zero to find the time it takes for the hammer and feather to fall to the ground. Solving \( 0 = -2.6t^2 + 4 \) results in \( 2.6t^2 = 4 \). Dividing both sides by 2.6, we get \( t^2 = \frac{4}{2.6} \). Finding the square root, we have \( t = \sqrt{\frac{4}{2.6}} \approx 1.24 \text{ sec} \).

Key Concepts

Acceleration in Vacuum

When objects are in a vacuum, they experience constant acceleration due to gravity, without the interference of air resistance. This is what was demonstrated by astronaut David Scott on the moon, where he dropped a hammer and a feather simultaneously. The absence of an atmosphere allowed them to fall with uniform acceleration despite different masses.

In mathematical terms, the constant acceleration in vacuum can be defined by a second-order differential equation. For example, the differential equation \( \frac{d^2 s}{d t^2} = -5.2 \text{ ft/sec}^2 \) signifies constant acceleration acting on the objects on the moon.

Important aspects of acceleration in vacuum include:

• Mass does not affect the rate of fall due to absence of air resistance.
• The acceleration remains constant, which simplifies calculations for velocity and displacement.
• Understanding this acceleration requires knowledge of differential equations to solve for velocity and position as functions of time.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation alongside given conditions that specify values at the initial point. In this context, when the hammer and feather were released, their initial velocity was \( \frac{d s}{d t} = 0 \) and they were 4 feet above the lunar surface, \( s = 4 \) when \( t = 0 \).

The process to approach such problems includes:

• Identifying the differential equation representing the motion; in this case, \( \frac{d^2 s}{d t^2} = -5.2 \).
• Using initial conditions to integrate the equation step-by-step; first finding the velocity, then the position as functions of time.
• Applying these conditions ensures you derive functions that accurately describe the system's behavior from the start.

By integrating twice starting from acceleration, both velocity and displacement functions can be obtained, ensuring the solution fits the initial scenario.

Integration in Calculus

Integration is a core concept in calculus used to find functions based on their rates of change. In the context of differential equations, like those describing a falling object, integration helps find velocity from acceleration and displacement from velocity.

The exercise uses integration as follows:

• First, integrate the acceleration \( \frac{d^2 s}{d t^2} = -5.2 \) to get the velocity function \( \frac{d s}{d t} = -5.2t + C_1 \).
• Apply the initial condition \( \frac{d s}{d t} = 0 \) when \( t = 0 \) to solve for \( C_1 \), leading to \( C_1 = 0 \).
• Next, integrate the velocity \( \frac{d s}{d t} = -5.2t \) to find the displacement function: \( s = -2.6t^2 + C_2 \).
• Use the condition \( s = 4 \) at \( t = 0 \) to solve for \( C_2 = 4 \). This results in \( s = -2.6t^2 + 4 \).

This two-step integration builds a complete picture of the motion, considering both velocity and distance over time.