Analysis of Uniformly Accelerated Motion

This page focuses on analyzing gleichmäßig beschleunigte Bewegung (uniformly accelerated motion) through graphs and problem-solving. It presents a velocity-time graph and a series of questions related to the motion of a stone launched vertically upward.

Example: A stone is launched vertically upward at t = 0 from a catapult located at x = 0. The velocity-time graph is provided, and students are asked to extract information and solve related problems.

The questions cover various aspects of motion analysis:

1. Determining initial velocity and acceleration from the graph
2. Writing the motion equations
3. Calculating the stone's position at different time intervals
4. Describing the motion in words
5. Finding specific positions (30 m above and below the catapult)

Highlight: The page emphasizes the importance of interpreting graphs and applying motion equations to solve real-world problems.

The solution process is detailed, showing step-by-step calculations for each part of the problem. This approach demonstrates how to use the Weg-Zeit-Gesetz gleichmäßig beschleunigte Bewegung (distance-time law for uniformly accelerated motion) and the Geschwindigkeit-Zeit-Gesetz (velocity-time law) in practical scenarios.

Graphical Representation and Problem Solving in Uniformly Accelerated Motion

This page continues the analysis of the stone's motion, focusing on graphical representation and problem-solving techniques for gleichmäßig beschleunigte Bewegung (uniformly accelerated motion).

The page presents a position-time graph (x-t diagram) based on calculations from the previous page. This graph visually represents the stone's motion over time.

Highlight: The position-time graph provides a clear visual representation of the stone's trajectory, showing its rise and fall.

A verbal description of the stone's motion is provided, explaining that it is launched with an initial velocity of 30 m/s, rises with decreasing velocity until it reaches its turning point at 6 seconds, and then falls back down with increasing velocity.

Example: The stone's motion is described as: "The stone is shot from the catapult at 30 m/s and then initially shoots upwards with ever-slowing velocity until the turning point, which it reaches after 6 seconds. From there it falls back down, and the velocity with which it falls becomes ever higher."

The final part of the page focuses on solving the problem of when the stone is 30 m above and below the catapult. This involves using the quadratic equation derived from the motion equation:

x(t) = -5m/s² · t² + 30m/s · t

The solution demonstrates the application of the quadratic formula to find the time values when the stone reaches these specific positions.

Vocabulary: The quadratic formula, also known as the "pq-Formel" in German, is used to solve quadratic equations in the form ax² + bx + c = 0.

This page effectively ties together the theoretical concepts of uniformly accelerated motion with practical problem-solving techniques, emphasizing the importance of both mathematical and graphical approaches in physics.