constant velocity particle model worksheet 2

Lemon

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01-02-2025

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This worksheet delves deeper into the constant velocity particle model, building upon your foundational understanding of motion. We'll explore more complex scenarios and problem-solving techniques. Remember, the constant velocity particle model simplifies objects as point particles moving at a consistent speed in a specific direction. This simplification allows us to focus on the core principles of kinematics.

Understanding Key Concepts

Before we tackle the problems, let's review some critical concepts:

• Displacement (Δx): The change in position of a particle. It's a vector quantity, meaning it has both magnitude (distance) and direction. Remember that displacement is not always equal to the distance traveled.
• Velocity (v): The rate of change of displacement. Like displacement, it's a vector quantity. Constant velocity implies both constant speed and constant direction.
• Time (t): The duration of the motion.
• The relationship between displacement, velocity, and time: The fundamental equation is: Δx = v * t

Tackling Different Scenarios

The constant velocity model is applicable in various situations, though real-world scenarios rarely exhibit perfectly constant velocity. However, the model serves as a valuable approximation in many cases. Let's consider some examples:

Scenario 1: Simple Linear Motion

A car travels at a constant velocity of 20 m/s east for 10 seconds. What is its displacement?

Solution:

Here, v = 20 m/s (east), and t = 10 s. Using the equation Δx = v * t, we get:

Δx = (20 m/s) * (10 s) = 200 m (east)

Scenario 2: Introducing Vectors

A bird flies at a constant velocity of 5 m/s at an angle of 30° north of east for 5 seconds. Calculate its displacement in both the x and y directions.

Solution:

This problem requires vector decomposition. We need to find the components of the velocity vector in the x (east) and y (north) directions.

• Vx = v * cos(30°) = 5 m/s * cos(30°) ≈ 4.33 m/s (east)
• Vy = v * sin(30°) = 5 m/s * sin(30°) = 2.5 m/s (north)

Now, we can calculate the displacement in each direction:

• Δx = Vx * t = 4.33 m/s * 5 s ≈ 21.65 m (east)
• Δy = Vy * t = 2.5 m/s * 5 s = 12.5 m (north)

Scenario 3: Relative Velocity

A boat travels across a river with a velocity of 10 m/s relative to the water. The river flows at a constant velocity of 5 m/s. What is the boat's velocity relative to the ground if it travels directly across the river? (Assume the river flow is perpendicular to the boat's direction)

Solution:

This problem involves vector addition. We can use the Pythagorean theorem to find the resultant velocity:

Resultant velocity = √(10² + 5²) = √125 ≈ 11.18 m/s

The direction can be found using trigonometry (arctan(5/10) = 26.6°). Therefore, the boat's velocity relative to the ground is approximately 11.18 m/s at an angle of 26.6° downstream from the direction directly across the river.

Practice Problems

Now, it's your turn to apply these concepts. Solve the following problems:

1. A train travels at a constant velocity of 60 km/h west for 30 minutes. What is its displacement?
2. A plane flies at a constant velocity of 800 km/h at an angle of 45° north of west for 2 hours. Calculate its displacement in both the x and y directions.
3. A cyclist rides at 15 m/s due north into a headwind of 5 m/s. What is the cyclist's velocity relative to the ground?

Remember to show your work and clearly state your units! Understanding the constant velocity model is crucial for building a strong foundation in physics. By mastering these fundamental principles, you'll be well-prepared for more complex concepts in future studies.