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Chapter 1: Problem 58

Emergency Landing. A plane leaves the airport in Galisteo and flies \(170 \mathrm{~km}\) at \(68.0^{\circ}\) east of north; then it changes direction to fly \(230 \mathrm{~km}\) at \(36.0^{\circ}\) south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

Short Answer

Expert verified
The rescue crew should fly in the north east direction. The exact direction and distance would be based on the calculations carried out in the steps outlined above.

Step by step solution

01

Convert into components

First, we need to convert the distances travelled into north-south and east-west components. For the first leg of the journey, the north component (y1) can be calculated using the formula \(y1 = 170 \,km * \cos(68.0^{\circ})\) and the east component (x1) can be calculated using the formula \(x1 = 170 \,km * \sin(68.0^{\circ})\). Similarly, for the second leg of the journey, the south component (y2) is \(y2 = 230 \,km * \sin(36.0^{\circ})\) and the east component (x2) is \(x2 = 230 \,km * \cos(36.0^{\circ})\)
02

Find total displacement

Next, we find the total displacement in the north-south direction (Y) and the east-west direction (X). The total north-south displacement (Y) is the north component of the first leg minus the south component of the second leg (\(Y = y1 - y2\)). The total east-west displacement (X) is the sum of the east components of the first and second legs (\(X = x1 + x2\))
03

Calculate distance and direction

Now that we have the total displacements in the north-south and east-west directions, we can find the total straight-line distance (D) from the airport to the plane using the Pythagorean theorem (\(D = \sqrt{X^2 + Y^2}\)). The direction of the rescue crew's flight (胃) can be found using the inverse tangent (\(胃 = arctan(\frac{Y}{X})\)). Since X and Y are both positive, this angle will be in the first quadrant, which means the rescue crew should fly northeast
04

Convert to degrees

Since we computed the direction in radians, it needs to be converted into degrees by multiplying with 180/蟺. This gives us the final direction for the rescue crew

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Key Concepts

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

Displacement
When we talk about displacement, we're describing the overall change in position of an object. It's different from distance, as it considers only the initial and final points, giving us both magnitude and direction.
For instance, if a plane travels a complex route but lands straight in front of its starting point, its displacement is zero.
In the context of flight, displacement helps us find the shortest path back鈥攊mportant during emergencies like the one in the scenario. We calculate displacement vectorially, often using components, to ensure we provide both how far and in which direction we need to proceed to aid the plane.
Components
Vectors have both magnitude and direction, and to work with them easily, we break them into components. These can be thought of as the vector's shadows on the x-axis and y-axis in a coordinate plane.
For the exercise, the plane's journey splits into north-south and east-west components. This breakdown helps in calculating precise path adjustments later on.
  • North-South component: aligns with vertical axis (y-axis).
  • East-West component: aligns with horizontal axis (x-axis).
Using trigonometric functions like sine and cosine, we derive these components from the initial angles and magnitudes, aiding in understanding the resultant navigation path.
Pythagorean Theorem
The Pythagorean theorem is essential for finding distances in right-angled triangles. Stated simply, it says the square of the hypotenuse (the triangle's longest side) equals the sum of the squares of the other two sides.
In vector addition, when we know the components, or sides of our right-angled triangle, we use the Pythagorean theorem to find the direct "as-the-crow-flies" distance.
In our scenario, once we found the composite north-south and east-west displacements, the theorem provides the straight-line distance from the airport to the plane's landing. This step is crucial in rescue missions, as it determines both the minimal distance and potential time to access the plane.
Trigonometry
Trigonometry bridges the gap between angles and side lengths in triangles. It uses functions like sine, cosine, and tangent to link an angle to ratios of sides.
For navigation and rescue operations, trigonometry is indispensable. It allows us to resolve an angled path into vertical and horizontal parts鈥攖he components.
  • Sine ( \( \sin \)) links an angle with the ratio of the opposite side to the hypotenuse.
  • Cosine ( \( \cos \)) relates the angle to the adjacent side and hypotenuse.
  • Tangent ( \( \tan \)) compares the opposite side directly to the adjacent side.
Our task uses the idea of the inverse tangent to determine the direction. If we have each side's length, trigonometry lets us find the angle the rescue crew must fly.

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Most popular questions from this chapter

Torque is a vector quantity that specifies the effectiveness of a force in causing the rotation of an object. The torque that a force \(\vec{F}\) exerts on a rigid object depends on the point where the force acts and on the location of the axis of rotation. If \(\vec{r}\) is the length vector from the axis to the point of application of the force, then the torque is \(\vec{r} \times \vec{F}\). If \(\vec{F}\) is \(22.0 \mathrm{~N}\) in the \(-y\) -direction and if \(\vec{r}\) is in the \(x y\) -plane at an angle of \(36^{\circ}\) from the \(+y\) -axis toward the \(-x\) -axis and has magnitude \(4.0 \mathrm{~m},\) what are the magnitude and direction of the torque exerted by \(\overrightarrow{\boldsymbol{F}} ?\)

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Find the angle between each of these pairs of vectors: (a) \(\vec{A}=-2.00 \hat{\imath}+6.00 \hat{\jmath}\) and \(\vec{B}=2.00 \hat{\imath}-3.00 \hat{\jmath}\) (b) \(\vec{A}=3.00 \hat{\imath}+5.00 \hat{\jmath} \quad\) and \(\quad \vec{B}=10.00 \hat{\imath}+6.00 \hat{\jmath}\) (c) \(\vec{A}=-4.00 \hat{\imath}+2.00 \hat{\jmath}\) and \(\vec{B}=7.00 \hat{\imath}+14.00 \hat{\jmath}\)

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