Problem 108 - Fundamentals of Physics

Question

A jumbo jet must reach a speed of \(360 \mathrm{~km} / \mathrm{h}\) on the runway for takcoff. What is the lowest constant accelcration needed for takcoff from a \(1.80 \mathrm{~km}\) runway?

Step-by-Step Solution

Answer

The lowest constant acceleration needed is approximately \(2.78 \text{ m/s}^2\).

Step 1: Understand the Problem

We need to find the lowest constant acceleration required for a jumbo jet to reach a takeoff speed of \(360 \text{ km/h}\) using a \(1.8 \text{ km}\) runway.

Step 2: Convert Units

Convert the speed from \(\text{km/h}\) to \(\text{m/s}\). Use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Therefore, \(360 \text{ km/h} = \frac{360}{3.6} \text{ m/s} = 100 \text{ m/s}\).

Step 3: Use the Kinematic Equation

Use the kinematic equation \(v^2 = u^2 + 2as\), where \(v\) is the final velocity (100 m/s), \(u\) is the initial velocity (0 m/s), \(a\) is the acceleration, and \(s\) is the distance (1800 m).

Step 4: Solve for Acceleration

Substitute the known values into the kinematic equation: \[100^2 = 0^2 + 2 \cdot a \cdot 1800\]Simplify and solve for \(a\):\[10000 = 3600a\]\[a = \frac{10000}{3600}\]\[a \approx 2.78 \text{ m/s}^2\]

Key Concepts

Acceleration

Acceleration is the rate at which an object's velocity changes over time. It is a vector, meaning it has both magnitude and direction. In the context of our jumbo jet problem, acceleration tells us how quickly the jet's speed increases as it travels down the runway to reach takeoff velocity.
We calculate acceleration by dividing the change in velocity by the time duration. However, when using kinematic equations, acceleration is directly derived from changes in speed and distance. It's crucial to remember that acceleration is expressed in meters per second squared (\( \text{m/s}^2 \)). In our example, the calculated acceleration was approximately \(2.78 \text{ m/s}^2 \), which means the jet's speed increases by \(2.78 \text{ m/s}\) for every second it travels along the runway.

Velocity Conversion

Velocity conversion is essential when working with different unit systems. The initial velocity given in the problem was in kilometers per hour (\( \text{km/h} \)). However, most kinematic equations require the velocity in meters per second (\( \text{m/s} \)).
To convert from \( \text{km/h} \) to \( \text{m/s} \), use the conversion factor: 1 \( \text{km/h} \) equals \(\frac{1}{3.6} \text{ m/s} \). So, a speed of \(360 \text{ km/h}\) is equivalent to \(100 \text{ m/s}\). This conversion is vital for consistency in calculations and understanding the problem in the context of kinematic equations.

Kinematic Equations

Kinematic equations are powerful tools in physics used to relate the motion of objects. They connect variables such as initial velocity, final velocity, acceleration, time, and displacement. In solving problems, these equations help predict one of these unknown values when others are known.
For the jumbo jet exercise, we used the equation: \[ v^2 = u^2 + 2as \] where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is acceleration, and \(s\) is the displacement. This equation allowed us to find the required acceleration without needing the time duration, focusing on the relation between speed change and distance covered.

Runway Distance

Runway distance is a critical factor in determining whether an aircraft can safely take off. It provides the space needed for the aircraft to accelerate to takeoff velocity. The longer the runway, the more time and space an aircraft has to achieve the necessary speed.
In our problem, the runway distance was \(1800 \text{ m}\). This distance is essential input for the kinematic equations to evaluate the acceleration required. It affects both the safety and feasibility of the takeoff process. Understanding how any changes in this variable might influence acceleration or speed requirements helps in making practical decisions in aviation.