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Magazine Chapter 3: Problem 8
(II) An airplane is traveling 835 km/h in a direction 41.5\(^\circ\) west of north (Fig. 3-34). (\(a\)) Find the components of the velocity vector in the northerly and westerly directions. (\(b\)) How far north and how far west has the plane traveled after 1.75 h?
Short Answer
Expert verified
North: 625.7 km/h, West: 553.9 km/h. After 1.75 h, North: 1095.0 km, West: 969.3 km.
Step by step solution
01
Understanding the Problem
We need to find the components of the velocity vector of an airplane traveling at 835 km/h, 41.5掳 west of north. Then, we will calculate how far it travels in the north and west directions after 1.75 hours.
02
Breaking Down the Velocity Vector
The velocity vector can be split into two perpendicular components: a northerly component and a westerly component. For a vector at an angle \( \theta \) from north, the northerly component \( v_{N} \) is found using the cosine function: \( v_{N} = v \cdot \cos(\theta) \), and the westerly component \( v_{W} \) using the sine function: \( v_{W} = v \cdot \sin(\theta) \).
03
Calculate the Northerly Velocity Component
Using the formula \( v_{N} = v \cdot \cos(\theta) \), where \( v = 835 \) km/h and \( \theta = 41.5^{\circ} \), calculate \( v_{N} = 835 \cdot \cos(41.5^{\circ}) \approx 625.7 \) km/h.
04
Calculate the Westerly Velocity Component
Use \( v_{W} = v \cdot \sin(\theta) \), so \( v_{W} = 835 \cdot \sin(41.5^{\circ}) \approx 553.9 \) km/h.
05
Calculate Distance Traveled North
Use the northerly velocity component to find the distance traveled north in 1.75 hours. The formula is \( \,\text{distance} = v_{N} \cdot \text{time} \): \( \,\text{distance north} = 625.7 \cdot 1.75 \approx 1095.0 \) km.
06
Calculate Distance Traveled West
Similarly, use the westerly velocity component to find the distance traveled west: \( \,\text{distance west} = 553.9 \cdot 1.75 \approx 969.3 \) km.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity Components
When discussing velocity in physics, a vector quantity, we often break it down into its components. This means separating it into parts that describe motion in specific directions. In this exercise, we deal specifically with breaking down the velocity of an airplane into two parts: a northerly component and a westerly component.
The concept of splitting a velocity into components enables us to understand and analyze motion separately along perpendicular directions, which is immensely useful in navigation and engineering applications.
The concept of splitting a velocity into components enables us to understand and analyze motion separately along perpendicular directions, which is immensely useful in navigation and engineering applications.
- **Northerly component** - Indicates how fast the object is moving northward.
- **Westerly component** - Indicates how fast the object is moving westward.
Northerly and Westerly Directions
In this specific problem, directions are crucial since they orient us regarding where the plane is headed. When we say 'northerly and westerly directions,' we are referring to the two perpendicular directions: northwards and westwards from a given point.
Geographically, north is typically considered the top of maps, with west to the left. So, when we say the plane is traveling 41.5掳 west of north, we imagine the angle formed between the northward line and the direction of travel, sweeping towards the west.
Geographically, north is typically considered the top of maps, with west to the left. So, when we say the plane is traveling 41.5掳 west of north, we imagine the angle formed between the northward line and the direction of travel, sweeping towards the west.
- The **angle west of north** describes how much the actual path deviate from directly heading north.
- Knowing the angle allows us to precisely calculate how much of the plane's speed contributes to moving north and how much contributes to moving west.
Trigonometric Functions
Trigonometric functions, specifically sine and cosine, are our go-to tools in physics when dealing with angles and right-angle triangles. They help in determining how much a vector leans towards one direction or another.
Sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)) originate from the realm of trigonometry, which studies relationships involving lengths and angles of triangles:
Sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)) originate from the realm of trigonometry, which studies relationships involving lengths and angles of triangles:
- **Cosine** - Helps us find the length of the adjacent side of the angle, which in this case, corresponds to the northerly component: \( v_{N} = v \cdot \cos(\theta) \).
- **Sine** - Assists in calculating the opposite side's length, pointing to the westerly component: \( v_{W} = v \cdot \sin(\theta) \).
Distance Calculation
Once we have calculated the velocity components, the next step is to use them for distance calculation. Distance is straightforwardly derived using the formula:\[ \text{distance} = \text{velocity} \times \text{time} \].
In our case, both the north and west distances can be computed using their respective velocity components and the time the plane travels, which is 1.75 hours in this situation.
In our case, both the north and west distances can be computed using their respective velocity components and the time the plane travels, which is 1.75 hours in this situation.
- **Distance North** - This is calculated using the northerly component \( v_{N} \): \( \text{distance north} = v_{N} \times 1.75 \).
- **Distance West** - Similarly, using the westerly component \( v_{W} \): \( \text{distance west} = v_{W} \times 1.75 \).
