Problem 2
Question
Two aircraft leave an airfield at the same time. One travels due north at an average speed of \(300 \mathrm{~km} / \mathrm{h}\) and the other due west at an average speed of \(220 \mathrm{~km} / \mathrm{h}\). Calculate their distance apart after fourhours.
Step-by-Step Solution
Verified Answer
The distance between the aircraft is approximately 1488.1 km.
1Step 1: Calculate Distance Travelled by Each Aircraft
To find out how far each aircraft has traveled, use the formula for distance, which is \( \text{Distance} = \text{Speed} \times \text{Time} \). For the northbound aircraft: \( 300 \text{ km/h} \times 4 \text{ hours} = 1200 \text{ km} \). For the westbound aircraft: \( 220 \text{ km/h} \times 4 \text{ hours} = 880 \text{ km} \).
2Step 2: Set Up the Right Triangle
Since one aircraft is traveling north and the other west, they form a right triangle with their paths. The distance the northbound aircraft traveled is one leg, and the distance the westbound aircraft traveled is the other leg.
3Step 3: Apply the Pythagorean Theorem
To find the hypotenuse of the right triangle, which is the distance between the two aircraft, use the Pythagorean theorem: \( c^2 = a^2 + b^2 \). Substitute \( a = 1200 \text{ km} \) and \( b = 880 \text{ km} \): \[ c^2 = 1200^2 + 880^2 \].
4Step 4: Calculate and Solve for c
Compute the squares: \( 1200^2 = 1440000 \) and \( 880^2 = 774400 \). Add these results: \( 1440000 + 774400 = 2214400 \). Take the square root of the sum to find \( c \): \( c = \sqrt{2214400} \approx 1488.1 \text{ km} \).
Key Concepts
Distance CalculationRight TriangleAverage SpeedMathematical Problem Solving
Distance Calculation
To calculate the distance traveled by an object, we can use a simple formula derived from basic kinematics. This formula is:
- Distance = Speed × Time
- The northbound aircraft covers: \(300 \text{ km/h} \times 4 \text{ hours} = 1200 \text{ km}\).
- The westbound aircraft covers: \(220 \text{ km/h} \times 4 \text{ hours} = 880 \text{ km}\).
Right Triangle
In mathematics, a right triangle is a triangle where one of the angles is exactly 90 degrees.
For such triangles, there is a special relationship between the lengths of the sides, known as the Pythagorean Theorem.
In this exercise, the path of each aircraft travels forms the two legs of a right triangle because they are moving perpendicularly: one north, the other west.
The two distances form the legs of the triangle:
- Northbound leg: 1200 km.
- Westbound leg: 880 km.
Average Speed
Average speed is a concept often used in physics and mathematics to determine how fast an object is moving over a certain period.
It is calculated by dividing the total distance traveled by the total time taken.
For constant speeds, like in the case of the aircraft, the average speed equals their constant traveling speed.
In this problem:
- The northbound aircraft's average speed is 300 km/h.
- The westbound aircraft's average speed is 220 km/h.
Mathematical Problem Solving
Mathematical problem solving involves breaking down a complex problem into more manageable parts, using systematic methods or mathematical principles.
In this exercise, solving the problem involves:
- Identifying the problem: Calculate the distance between the two aircraft.
- Breaking it into smaller parts: Firstly, calculate the distance covered by each aircraft.
- Applying mathematical concepts: Use the Pythagorean Theorem to calculate the direct distance, i.e., the hypotenuse of the triangle formed.
- Executing calculations: Compute squares and square roots to solve for the distance.
Other exercises in this chapter
Problem 3
If \(\cos X=\frac{9}{41}\) determine the value of the other five trigonometry ratios.
View solution Problem 5
Point \(A\) lies at co-ordinate \((2,3)\) and point \(B\) at \((8,7)\). Determine (a) the distance \(A B\), (b) the gradient of the straight line \(A B\), and (
View solution Problem 8
Evaluate, correct to 4 decimal places: (a) sine \(168^{\circ} 14^{\prime}\) (b) cosine \(271.41^{\circ}\) (c) tangent \(98^{\circ} 4^{\prime}\)
View solution