/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A rocket is launched into the ai... [FREE SOLUTION] | 91Ó°ÊÓ

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A rocket is launched into the air. A few moments after liftoff, the rocket is 40 \(\mathrm{m}\) above the ground. After another 5 \(\mathrm{s}\) , the rocket is now 200 \(\mathrm{m}\) off the ground. What is the average velocity of the rocket during the 5 s part of the flight? (A) 16 \(\mathrm{m} / \mathrm{s}\) (B) 32 \(\mathrm{m} / \mathrm{s}\) (C) 48 \(\mathrm{m} / \mathrm{s}\) (D) 64 \(\mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
The average velocity of the rocket during the 5 s part of the flight was 32 \(\mathrm{m} / \mathrm{s}\).

Step by step solution

01

Determine the change in distance

First, we need to determine how much the rocket has moved in the given time. It moved from 40 m to 200 m, so the change in distance \(\Delta d\), is \(200 m - 40 m = 160 m\).
02

Determine the change in time

The change in time, \( \Delta t \), is given as 5 seconds.
03

Calculate the average velocity

Plugging our values into the average velocity formula, we get \(\frac{\Delta d}{\Delta t} = \frac{160 m}{5 s} = 32 m/s\). So the average velocity of the rocket during these 5 seconds was 32 m/s.

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Key Concepts

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

Average Velocity Formula
Understanding the average velocity of an object involves calculating the overall change in position (or displacement) with respect to the total time taken. This is expressed in the average velocity formula:
\[\begin{equation}\text{Average velocity} = \frac{\Delta \text{position}}{\Delta \text{time}}\end{equation}\]
In the context of the rocket problem, the rocket's change in position, often referred to as the displacement (\Delta d), is the difference between its final and initial positions. The change in time(\Delta t), is the duration the rocket took to move between these two points. The average velocity can be both positive and negative, indicating the direction of the motion relative to a chosen reference point. It's important to note that average velocity is different from average speed, as velocity includes both magnitude and direction, while speed only considers magnitude.
Kinematics
Kinematics is the branch of classical mechanics that describes the motion of points, objects, and systems of bodies without considering the forces that cause them to move. It provides a framework for analyzing motion using concepts such as displacement, velocity, and acceleration, among others.
Within kinematics, average velocity is a fundamental concept, applying to various problems, from simple ones such as our rocket's flight to complex scenarios involving multiple forces and directions. In essence, kinematics simplifies the physics of motion by reducing it to its most basic elements, allowing us to calculate an object's position in space over time.
Physics Problem Solving
Solving physics problems often involves a systematic approach, breaking down complex phenomena into solvable steps. The key to effective problem-solving in physics lies in understanding the core concepts and their formulaic representations.
Our rocket problem showcases a typical process:
  • Identify the given information and what you need to find.
  • Determine the relevant physics principles involved, such as the average velocity formula in this case.
  • Perform the necessary computations by substituting known values into the formulas.
  • Interpret the results within the context of the problem.

The rocket problem required us to first determine the displacement and the time interval, which then allowed us to calculate the average velocity using the appropriate formula. This step-by-step method ensures a strategic and methodical path to the solution, which can be applied to a wide array of physics problems.

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Most popular questions from this chapter

Questions 1-3 refer to the following scenario: IMAGE IS NOT AVAILABLE TO COPY An explorer travels 30 \(\mathrm{m}\) east, then 20\(\sqrt{2} \mathrm{m}\) in a direction \(45^{\circ}\) south of east, and then 140 \(\mathrm{m}\) north. What is the distance traveled by the explorer? (A) 167.2 \(\mathrm{m}\) (B) 169 \(\mathrm{m}\) (C) 170 \(\mathrm{m}\) (D) 198.2 \(\mathrm{m}\)

Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free- response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part. (DIAGRAM IS NOT AVAILABLE TO COPY) A machine launches a 2 \(\mathrm{kg}\) ball to the right with an initial velocity 16 \(\mathrm{m} / \mathrm{s}\) at a launched angle of \(30^{\circ}\) to a student standing 20 \(\mathrm{m}\) away with a baseball bat. (DIAGRAM IS NOT AVAILABLE TO COPY) (a) What height must the student swing the bat to hit the ball? (b) What is the magnitude of the velocity of the ball just before impact? (c) If the student hits the ball with an upwards vertical velocity of 5 h/s and horizontal velocity of 12 \(\mathrm{m} / \mathrm{s}\) to the left, what are the horizontal and vertical components of the impulse of the ball from the collision? (d) If the impact time with the bat was 0.05 s, what is the magnitude of the average force experienced by the ball during impact? (e) What is the direction of the average force experienced by the ball during impact?

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A horizontal spring is attached to a 5 kg block. When the block is pulled 5 cm to the right, the restoring force has a magnitude of 6 N. What is the frequency of the spring? (A) 0.32 Hz (B) 0.56 Hz (C) 0.78 Hz (D) 0.98 Hz

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