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Displacement is a term that describes how far an object has moved from its initial position. It is a vector quantity, meaning it has both magnitude and direction. In the context of this rocket exercise, displacement is essentially the change in the rocket's position from its starting point. Consider the rocket's motion in part (a) of the problem. Initially, the rocket is at 63 meters above the ground. After 4.75 seconds, it reaches a height of 1000 meters. The displacement here is the difference between the two positions: - Initial position: 63 meters - Final position: 1000 meters Hence, the displacement \( \Delta s \) is calculated as follows: \[ \Delta s = 1000 \, \text{m} - 63 \, \text{m} = 937 \, \text{m} \]In part (b), the displacement is the rocket's position from lift-off, which is 0 meters, to when it reaches 1000 meters at 5.90 seconds. Therefore: \[ \Delta s = 1000 \, \text{m} - 0 \, \text{m} = 1000 \, \text{m} \]

The time interval in a physics problem is crucial to determining how to calculate the velocity or any rate-related value. It represents the duration over which a certain change has occurred. In this rocket problem, two time intervals are of concern: - For part (a), the time interval is the duration from when the rocket reaches the top of the launch platform (1.15 s after lift-off) to when it is 1.00 km above the ground. This interval lasts 4.75 seconds. - For part (b), the time interval spans from the lift-off (0 seconds) to when it is 1.00 km above the ground, which happens at 5.90 seconds. By understanding these, we can apply them to the average velocity calculation: - Part (a) time interval: 4.75 s - Part (b) time interval: 5.90 s

In this exercise, we delve into the exciting world of rocket motion. Rocket motion involves rapid acceleration and significant changes in position over short periods. Understanding how a rocket moves can offer insight into the fundamental principles of physics. Rockets operate by expelling high-speed exhaust gases, propelling them forward. In our exercise, the rocket is moving upwards from the moment of lift-off. We look at its height changes over certain time periods. - Upon lift-off, the rocket accelerates, moving upwards; we observe its motion first by the height it clears at the launch platform (63 m) in 1.15 seconds. - After a total of 5.90 seconds, the rocket reaches 1.00 km in height. These distances allow us to calculate its average velocity using its motion data. Rocket motion is constant in terms of its directional change from ground upwards, and it aids any velocity or displacement calculations by setting the framework within which they are calculated.

Understanding how to calculate velocity is essential in physics, particularly for analyzing motion. Average velocity is a common way to understand speed over a period.The formula for average velocity is:\[ v_{avg} = \frac{\Delta s}{\Delta t} \]Here, \( \Delta s \) represents displacement, and \( \Delta t \) is the time interval.Let's see how this applies to our rocket problem:- **Part (a):** - Displacement: 937 m - Time Interval: 4.75 s - Average Velocity: \[ v_{avg} = \frac{937 \, \text{m}}{4.75 \, \text{s}} \approx 197.26 \, \text{m/s} \]- **Part (b):** - Displacement: 1000 m - Time Interval: 5.90 s - Average Velocity: \[ v_{avg} = \frac{1000 \, \text{m}}{5.90 \, \text{s}} \approx 169.49 \, \text{m/s} \]These calculations demonstrate how displacement and elapsed time come together to define how fast an object like a rocket is moving, averaged over a specified time. This method helps in understanding the rocket's behavior as it moves through its flight path.