Problem 92 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A pilot lands a fighter aircraft on an aircraft carrier. At the moment of touchdown, the speed of the aircraft is \(160 \mathrm{mph}\). If the aircraft is brought to a complete stop in 1 sec and the deceleration is assumed to be constant, find the number of \(g\) 's the pilot is subjected to during landing \(\left(1 \mathrm{~g}=32 \mathrm{ft} / \mathrm{sec}^{2}\right) .\)

Step-by-Step Solution

Answer

The pilot is subjected to \(\text{Number of g's} = \frac{\left[\frac{0 - (160 \times 5280 \div 3600)}{1}\right]}{32} \approx 6.25 ~g\) during landing.

Step 1: Convert speed to ft/sec

First, we need to convert the speed of the aircraft from miles per hour to feet per second: \[160 \frac{\text{miles}}{\text{hour}} * \frac{5280 ~\text{feet}}{1 ~\text{mile}} * \frac{1 ~\text{hour}}{3600 ~\text{seconds}}\]

Step 2: Calculate the deceleration rate

Now, we use the given information that the aircraft comes to a complete stop in 1 second and the deceleration is constant. We can use the formula \(v_f = v_i + at\), where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration (or deceleration in this case), and \(t\) is the time. Since our goal is to find the deceleration rate, we can re-write the equation as: \[a = \frac{v_f - v_i}{t}\]

Step 3: Plug in the values and solve for deceleration

The final velocity (\(v_f\)) is 0 (as it comes to a stop), the initial velocity (\(v_i\)) is the result from Step 1 (in ft/sec), and the time (\(t\)) is 1 second. Plugging in these values and solving for the deceleration rate: \[a = \frac{0 - (160 \times 5280 \div 3600)}{1}\]

Step 4: Calculate the number of g's the pilot is subjected to

Now that we have found the deceleration rate (in ft/sec^2), we can divide it by the standard acceleration due to gravity (32 ft/sec^2) to find the number of g's the pilot is subjected to: \[\text{Number of g's} = \frac{a}{32 \frac{\text{ft}}{\text{sec}^2}}\]

Step 5: Simplify and find the result

Now, calculate the result by plugging in the values from step 3, and simplifying the expression: \[\text{Number of g's} = \frac{\left[\frac{0 - (160 \times 5280 \div 3600)}{1}\right]}{32}\] After solving this, we can find the number of g's the pilot is subjected to during the landing process.

Key Concepts

Deceleration

In simple terms, deceleration is the process of slowing down. When an object reduces its speed, it undergoes deceleration, which is a type of acceleration but in the opposite direction.
For the aircraft landing on the carrier, deceleration means it is reducing its speed to come to a complete stop.

• During the touchdown, the aircraft moves from a high speed to zero.
• Deceleration is calculated using the formula: \[ a = \frac{v_f - v_i}{t} \] where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is time.
• Here, \( v_f \) is 0 because the aircraft stops completely.

This formula helps us find the rate at which the speed of the aircraft decreases. In this scenario, the fast reduction in speed subjects the pilot to forces measured in g's, a unit explaining how strong these forces are compared to gravity.

Conversions

Conversions play a critical role when working with different measurement units. They allow us to understand and manipulate quantities in a standardized way. Let's consider converting speed in this scenario as an example.
Imagine the aircraft's speed initially given in miles per hour (mph).

• Miles per hour doesn't suit physics equations that often use feet per second.
• To convert 160 mph to feet per second, use: \[ 160 \frac{\text{miles}}{\text{hour}} \times \frac{5280 ~\text{feet}}{1 ~\text{mile}} \times \frac{1 ~\text{hour}}{3600 ~\text{seconds}} \]

This conversion provides a clear understanding by breaking down the larger units, making the deceleration calculation straightforward and precise.

Physics Concepts

Physics helps us comprehend how objects move and interact. In this exercise, several physics concepts are at play, enhancing our understanding of the aircraft's landing process and the pilot's experience.
Let's dive into these underlying principles:

• Velocity and Acceleration: Velocity is the speed in a specific direction. When the aircraft stops, the change in velocity per time unit is known as acceleration, or deceleration when slowing down.
• Impact of Gravity: Forces during landing feel like multiple g's, referring to multiples of Earth's gravitational pull \( 1 \text{ g} = 32 \text{ ft/sec}^2 \).
• Force on Pilot: When the aircraft stops suddenly, inertial forces act on the pilot. These forces are critical in designing safety systems.

Understanding these physics concepts is key. They reveal how everyday phenomena, like a plane landing, are governed by fundamental laws of motion and force.