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Chapter 3: Problem 57

You are in a hot-air balloon that, relative to the ground, has a velocity of 6.0 m/s in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is 2.0 m/s. What are the magnitude and direction of the hawk鈥檚 velocity relative to the ground? Express the directional angle relative to due east.

Short Answer

Expert verified
The hawk's velocity is 6.32 m/s at 18.43掳 north of east relative to the ground.

Step by step solution

01

Understand the problem

We have two velocity vectors: the hot-air balloon moving east at 6.0 m/s and the hawk moving north at 2.0 m/s relative to the balloon. We need to find the hawk's velocity relative to the ground.
02

Set up the reference frames

Set the hot-air balloon's velocity as the reference: \[\vec{v}_{bg} = 6.0 \text{ m/s (east)}\].The hawk's velocity relative to the balloon: \[\vec{v}_{hb} = 2.0 \text{ m/s (north)}\].We need to find \(\vec{v}_{hg}\) which is the hawk's velocity relative to the ground.
03

Use vector addition

The velocity of the hawk relative to the ground is the vector sum of the balloon's velocity relative to the ground and the hawk's velocity relative to the balloon:\[\vec{v}_{hg} = \vec{v}_{bg} + \vec{v}_{hb}\].Thus,\[\vec{v}_{hg} = 6.0 \text{ m/s (east)} + 2.0 \text{ m/s (north)}\].
04

Calculate the magnitude of the velocity

Use the Pythagorean theorem to calculate the magnitude of \(\vec{v}_{hg}\):\[\|\vec{v}_{hg}\| = \sqrt{(6.0)^2 + (2.0)^2} = \sqrt{36 + 4} = \sqrt{40} = 6.32 \text{ m/s}\].
05

Determine the direction

Determine the direction using the tangent function:\[\theta = \tan^{-1}\left(\frac{2.0}{6.0}\right)\].Calculate \(\theta\) using a calculator:\[\theta = \tan^{-1}\left(\frac{2.0}{6.0}\right) \approx 18.43^\circ\].The direction is \(18.43^\circ\) north of east.

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Key Concepts

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

Vector Addition
In problems involving motion, particularly with multiple objects, vector addition is crucial to understand. Here, you are given two velocity vectors: the balloon moving east and the hawk moving north, but the hawk's northward velocity is given relative to the balloon. To find a vector that represents the hawk's true path relative to the ground, we sum these separate velocities.
  • The balloon's eastward velocity is expressed as a vector pointing straight east with a magnitude of 6.0 m/s.
  • The hawk's northward velocity, relative to the balloon, is a vector pointing north with a 2.0 m/s magnitude.
By adding these vectors - using vector addition - we find the true velocity of the hawk relative to the ground. Vector addition essentially combines these two directional velocities into one resultant vector.
Pythagorean Theorem
The Pythagorean theorem helps us solve the problem by calculating the magnitude of the resultant velocity vector. Suppose you consider the eastward and northward velocities as the two sides of a right-angled triangle.
  • The eastward velocity (6.0 m/s) is one side.
  • The northward velocity (2.0 m/s) is the other side.
The resultant vector that combines these forms the hypotenuse of the triangle. Using the formula \[\text{Hypotenuse} = \sqrt{(\text{East Velocity})^2 + (\text{North Velocity})^2}\]we get:\[\|\vec{v}_{hg}\| = \sqrt{(6.0)^2 + (2.0)^2} = 6.32 \text{ m/s}.\] This gives you the magnitude of the hawk's velocity relative to the ground.
Tangent Function
Finding the directional angle involves trigonometric functions, specifically the tangent function. Tangent relates the angle of a right triangle to the ratio of the opposite side to the adjacent side.To find the angle the velocity vector makes with the positive x-axis (east direction), use:\[\theta = \tan^{-1}\left(\frac{\text{North Velocity}}{\text{East Velocity}}\right)\]Here, substituting the values:\[\theta = \tan^{-1}\left(\frac{2.0}{6.0}\right) \approx 18.43^\circ\]This angle represents how far north of east the hawk is heading, relative to the ground.
Velocity Magnitude
The magnitude of velocity tells you how fast something is moving, regardless of direction. After using vector addition to find the resultant vector, the calculation of magnitude becomes a way to find the speed relative to the ground. In this context, it means: - The speed at which the hawk appears to be moving from a stationary point on the ground. - Calculated using the distance in the xy-plane (using Pythagorean theorem) traveled per unit of time. This highlights the importance of velocity vectors' magnitude for understanding real-world motion, giving a fuller picture beyond just direction.
Directional Angle
Directional angles are essential when talking about vectors as they place vectors precisely on a plane. For this problem, the angle relative to due east clarifies the actual path of the hawk from a compass perspective: - Measured counterclockwise from the east (x-axis). - The clockwise convention establishes consistency in measuring angles, making understanding directions intuitive, whether solving a math problem or navigating in real life.

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Most popular questions from this chapter

SSM A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.870 m/s at an angle of 35.0 above the table, and it lands on the magazine 0.0770 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.

SSM A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 162 km away. The radar antenna is pointing upward at an angle of 62.3 from the ground. Find the x and y components (in km) of the position vector of the satellite, relative to the antenna.

A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of \(2.40 \times 10^{2} {m} / {s}\) . The plane is to make a roundtrip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a 57.8 -m/s wind from the jet stream, which blows from west to east. What is the maximum distance that the plane can travel due west and just be able to return home?

Multiple-Concept Example 4 provides useful background for this problem. A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before striking the water?

In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in Figure 3.10 and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?
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