Problem 31 - University Calculus: Early Transcendentals

Question

Suppose that the distance an aircraft travels along a runway before takeoff is given by \(D=(10 / 9) t^{2},\) where \(D\) is measured in meters from the starting point and \(t\) is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches \(200 \mathrm{km} / \mathrm{h}\). How long will it take to become airborne, and what distance will it travel in that time?

Step-by-Step Solution

Answer

The aircraft takes 25 seconds to become airborne and travels approximately 694.44 meters in that time.

Step 1: Convert speed from km/h to m/s

The aircraft needs to reach a speed of 200 km/h to become airborne. First, convert this speed to meters per second by using the conversion 1 km/h = 1000 meters/3600 seconds: \[ 200 \text{ km/h} = \frac{200 \times 1000}{3600} \text{ m/s} = \frac{500}{9} \text{ m/s} \]

Step 2: Find the time using velocity formula

The velocity, the rate of change of distance with respect to time, is the derivative of the distance function. Given that \(D(t) = \frac{10}{9} t^2\), differentiate this with respect to time \(t\) to find the velocity function: \[ v(t) = \frac{dD}{dt} = \frac{20}{9} t \]Set this equal to the required speed and solve for \(t\): \[ \frac{20}{9} t = \frac{500}{9} \]\[ t = \frac{500}{20} = 25 \text{ seconds} \]

Step 3: Calculate the distance at takeoff time

Substitute the time \(t = 25\) seconds into the original distance function to find the distance the aircraft has traveled: \[ D(25) = \frac{10}{9} \times 25^2 = \frac{10}{9} \times 625 = \frac{6250}{9} \approx 694.44 \text{ meters} \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how things change. By using differentiation, we can find how a function changes its value as we tweak its input values. It's like figuring out the slope of a hill, depending on where you stand. In this exercise, we need to determine how fast an aircraft's distance changes over time. Mathematically, this involves finding the derivative of the distance function with respect to time.

To do this, we start with the distance function given by:

• \( D(t) = \frac{10}{9}t^2 \)

The derivative of this function, denoted as \( \frac{dD}{dt} \), represents the velocity or rate at which the distance is changing. Differentiating \( D(t) \) with respect to \( t \) gives us the velocity function:

• \( v(t) = \frac{20}{9} t \)

This velocity function tells us that the aircraft's speed increases linearly with time. By understanding differentiation, we can predict how quickly the aircraft accelerates along the runway.

Velocity

Velocity is a measure of how fast something is moving in a particular direction. In this context, it tells us how quickly the aircraft is advancing along the runway. Essentially, velocity is the derivative of the position function with respect to time. Therefore, it is expressed as:

• \( v(t) = \frac{dD}{dt} \)

This concept allows us to move from knowing where something is at a given time, to understanding how quickly it moves from one position to another over time.

In the problem, we are given a target velocity of 200 km/h, which we convert to meters per second as follows:

• \( 200 \text{ km/h} = \frac{500}{9} \text{ m/s} \)

With the velocity function derived from differentiation, \( v(t) = \frac{20}{9}t \), we set this equal to our target speed and solve for \( t \) to find out when the aircraft reaches this velocity, leading to:

• \( t = 25 \text{ seconds} \)

This result tells us how long it takes for the aircraft to reach the necessary takeoff speed.

Distance-Time Relationship

The distance-time relationship describes how an object's position changes as time progresses. Understanding this relationship is crucial for problems involving motion and change over time. In our example, this relationship is captured by the equation:

• \( D(t) = \frac{10}{9}t^2 \)

This equation tells us that distance increases quadratically with time. This means the longer the aircraft rolls on the runway, the distance covered grows more rapidly, as the square of time. Such relationships are important when examining accelerated motion as seen in take-off situations.

In the problem, once we know how long the aircraft travels before takeoff, determined by when the speed equals \( \frac{500}{9} \text{ m/s} \) at 25 seconds, we can calculate the precise distance traveled using the distance function:

• \( D(25) = \frac{6250}{9} \approx 694.44 \text{ meters} \)

This calculation helps us understand how far the aircraft moves before it becomes airborne, linking the equation of motion with real-world distances.