Airplane collision. An F-16 jet fighter is leaving from Rygge airfield, which we use as the origin of our coordinate system, at \(t=0.0 \mathrm{~s}\), and travels with a constant velocity \(\mathbf{v}_{1}=1700.0 \mathrm{~km} / \mathrm{h} \mathbf{j}\) towards the North. At the same time, an Airbus \(\mathrm{A} 310\) airplane is passing over Oslo, which is located at \(\mathbf{r}_{1}=-10 \mathrm{~km} \mathbf{i}+80 \mathrm{~km} \mathbf{j}\). The Airbus travels with a constant velocity of \(\mathbf{v}_{2}=105 \mathrm{~km} / \mathrm{h} \mathbf{i}+905 \mathrm{~km} / \mathrm{h} \mathbf{j}\). They are both travelling at the same height. (a) Find the position of the jet fighter as a function of time. (b) Find the position of the Airbus as a function of time. (c) Sketch the trajectories of both planes in the same diagram. (You can do this on your computer if you like.) (d) Will the airplanes collide? (e) If the airplanes are within a distance of \(1 \mathrm{~km}\) of each other, an alarm will sound in the plane, and an evasive maneuver will be attempted. Will the planes pass that close to each other?

Step by step solution

01

Position of the jet fighter

The jet fighter is moving with a constant velocity \(\mathbf{v}_{1} = 1700.0 \, \mathrm{km/h} \, \mathbf{j}\). To find the position of the jet fighter as a function of time, use the equation \(\mathbf{r}_{\mathrm{jet}}(t) = \mathbf{r}_{0} + \mathbf{v}_{1} t\). Here, the initial position \(\mathbf{r}_{0} = \mathbf{0}\), making the position function: \(\mathbf{r}_{\mathrm{jet}}(t) = 1700.0 \, t \, \mathbf{j}\).

02

Position of the Airbus

The Airbus starts from the point \(\mathbf{r}_{1} = -10 \, \mathrm{km} \, \mathbf{i} + 80 \, \mathrm{km} \, \mathbf{j}\) and has a constant velocity \(\mathbf{v}_{2} = 105 \, \mathrm{km/h} \, \mathbf{i} + 905 \, \mathrm{km/h} \, \mathbf{j}\). Use the equation \(\mathbf{r}_{\mathrm{Airbus}}(t) = \mathbf{r}_{1} + \mathbf{v}_{2} t\), resulting in the position function: \(\mathbf{r}_{\mathrm{Airbus}}(t) = (-10 + 105 t) \, \mathbf{i} + (80 + 905 t) \, \mathbf{j}\).

03

Sketch the trajectories

In a 2D coordinate system, sketch the paths of both planes. For the jet fighter, draw a straight line starting from the origin and moving upwards (along the y-axis) with a slope equivalent to a speed of 1700 km/h. For the Airbus, start from the point (-10, 80) and draw a line moving up and to the right with a slope determined by the speed components (105, 905) km/h.

04

Check for collision

Set the position functions equal to each other and solve for time to check for a potential collision. This results in \(1700 t \, \mathbf{j} = (-10 + 105 t) \, \mathbf{i} + (80 + 905 t) \, \mathbf{j}\). Equate \(\mathbf{i}\) and \(\mathbf{j}\) components separately:\ \(1700 t = 80 + 905 t\) and \(0 = -10 + 105 t\). The second equation gives \(t = \frac{10}{105} \approx 0.095 \, h\), solving for \$t\$ in the first equation: \(1700 t = 80 + 905 t, \ Rightarrow 795 t = 80, \ Rightarrow t \approx 0.101 \, h\). Because the times contradict, there is no collision.

05

Check for minimum distance within 1 km

Calculate the distance between the planes at each time step to ensure accuracy within 1 km. Use the position functions: \(\| \mathbf{r}_{\mathrm{jet}}(t) - \mathbf{r}_{\mathrm{Airbus}}(t) \| = \sqrt{(10 - 105 t)^2 + (1700 t - 80 - 905 t)^2} \). Check the minimum value of this distance function, solving for times around the point calculated previously.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

constant velocity

In physics, constant velocity means that an object's speed and direction remain unchanged over time. This concept is fundamental because it simplifies calculations and predictions. For the jet fighter in the exercise, the constant velocity is given as \( \textbf{v}_{1} = 1700 \text{ km/h} \text{ j} \) . Here's what this means:

• Speed: The jet fighter is traveling at 1700 km/h.
• Direction: The 'j' unit vector indicates it's moving directly north.

Knowing the constant velocity, we can determine its position at any time using the equation for linear motion.

position function

The position function describes an object's location at any given time. We derive it by integrating the velocity function over time. For the jet fighter, we start from the origin \( \textbf{r}_{0} = \textbf{0} \) and use its constant velocity:

The position function of the jet fighter becomes: \[ \textbf{r}_{\text{jet}}(t) = \textbf{v}_{1} t + \textbf{r}_{0} = 1700t \textbf{j} \] Thus, at any time \( t \), the jet's position is north of the start by \( 1700t \) km.
For the Airbus, it starts at \( \textbf{r}_{1} = -10 \textbf{i} + 80 \textbf{j}\text{k} \) and moves with velocity \( \textbf{v}_{2} = 105 \textbf{i} + 905 \textbf(j) \). So the position function becomes:
\[ \textbf{r}_{\text{Airbus}}(t) = \textbf{r}_{1} + \textbf{v}_{2} t = (-10 + 105t) \textbf{i} + (80 + 905t) \textbf{j} \]

trajectory sketching

Trajectory sketching involves plotting the paths of moving objects over time. This visual tool makes understanding their motion easier.

• Jet Fighter: Starts at the origin \( (0, 0) \) and moves straight up the y-axis due to its northward constant velocity. The trajectory is a line with a slope equal to 1700 km/h.
• Airbus: Begins at \( (-10, 80) \) km and moves up and to the right. Its velocity components determine the slope. Combining horizontal motion at 105 km/h and vertical motion at 905 km/h forms a steeper line.

Sketch both lines on a graph to visualize their paths, ensuring they represent the right velocities and starting points.

collision detection

To detect a collision, we need to see if the planes occupy the same position at any time.
Set their position functions equal: \[ 1700t \textbf{j} = (-10 + 105t) \textbf{i} + (80 + 905t) \textbf{j} \] Separate into horizontal and vertical components:

• \textbf{Horizontal (i):} \( 0 = -10 + 105t \rightarrow t = \frac{10}{105} \text{ h} \)
• \textbf{Vertical (j):} \( 1700t = 80 + 905t \rightarrow t = \frac{80}{795} \text{ h} \)

Since the times differ (approximately 0.095 h and 0.101 h), the planes won't collide. They don't share the same position at any simultaneous moment.

minimum distance

Even if there is no collision, we must check if the planes get dangerously close, within 1 km.
The distance function measures the gap between positions over time: \[ \textbf{d}(t) = \textbf{r}_{\text{jet}}(t) - \textbf{r}_{\text{Airbus}}(t) = [10 - 105t] \textbf{i} + [1700t - (80 + 905t)] \textbf{j} \] Simplify to get: \[ \textbf{d}(t) = (10 - 105t) * \textbf{i} + (795t - 80) * \textbf{j} \] Calculate the magnitude \( \textbf{d}_\text{mag}(t) \): \[ \textbf{d}_{\text{mag}}(t) = \sqrt{(10 - 105t)^2 + (795t - 80)^2} \] Evaluate this equation for critical points, especially around previously calculated times. This ensures we catch the minimum distance and check if it breaches the 1 km threshold. Calculating this correctly prevents false collision alerts.