91Ó°ÊÓ

A faulty model rocket moves in the \(x y\) -plane (the positive \(y\) -direction is vertically upward). The rocket's acceleration has components \(a_{x}(t)=\alpha t^{2}\) and \(a_{y}(t)=\beta-\gamma t,\) where \(\alpha=2.50 \mathrm{~m} / \mathrm{s}^{4}\) \(\beta=9.00 \mathrm{~m} / \mathrm{s}^{2},\) and \(\gamma=1.40 \mathrm{~m} / \mathrm{s}^{3} .\) At \(t=0\) the rocket is at the origin and has velocity \(\overrightarrow{\boldsymbol{v}}_{0}=v_{0 x} \hat{\imath}+v_{0 y} \hat{\jmath}\) with \(v_{0 x}=1.00 \mathrm{~m} / \mathrm{s}\) and \(v_{0 y}=7.00 \mathrm{~m} / \mathrm{s} .\) (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) What is the horizontal displacement of the rocket when it returns to \(y=0 ?\)

Step by step solution

01

Determine the velocity vectors

The velocity as a function of time can be obtained by integrating the acceleration functions. For \( v_{x}(t) \), the velocity in the x-direction, integrate the acceleration \( a_{x}(t) = \alpha t^{2} \) from 0 to t and add the initial velocity \( v_{0x} \) to it. For \( v_{y}(t) \), the velocity in the y-direction, integrate the acceleration \( a_{y}(t) = \beta - \gamma t \) from 0 to t and add the initial velocity \( v_{0y} \) to it.

02

Compute the position vectors

Similar to velocity, the position as a function of time is obtained by integrating the velocity functions from step 1. For \( x(t) \), the position in the x-direction, integrate the velocity function \( v_{x}(t) \) from 0 to t. For \( y(t) \), the position in the y-direction, integrate the velocity function \( v_{y}(t) \) from 0 to t.

03

Determine the maximum height reached by the rocket

The maximum height reached by the rocket occurs when the vertical velocity becomes zero, i.e., when \( v_{y}(t) = 0 \). Set the velocity function \( v_{y}(t) \) equal to zero and solve for \( t \). Substituting this \( t \) value into the position function \( y(t) \) will yield the maximum height.

04

Calculate the horizontal displacement when the rocket returns to y=0

The horizontal displacement of the rocket when it returns to ground, i.e., when \( y(t) = 0 \), can be found by setting the position function \( y(t) \) equal to zero. Solve for \( t \) and substitute it into the position function \( x(t) \) to find the horizontal displacement at that time.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinematics

Understanding kinematics is crucial for analyzing the motion of objects without considering the forces causing the motion. In kinematics, we study quantities like velocity, acceleration, displacement, and time. These elements help us describe the trajectory of objects, such as rockets traveling through an xy-plane.
For projectile motion, the motion can be split into horizontal and vertical components. Each component acts independently from the other.
A great way to grasp kinematics in the context of our exercise is to focus on the problem's breakdown:

• Identify initial conditions: The rocket's initial position and its initial velocity components.
• Recognize motion functions: The given functions for acceleration need to be applied to derive velocity and position over time.
• Analyze separate dimensions: The x- and y- components are treated independently when integrating to get velocity and position functions.

Breaking down these components helps in computing parameters like maximum height and horizontal displacement.

Acceleration Components

The acceleration of the rocket has components in both the x and y directions. In this exercise, we see specific functions representing these components:

• In the x-direction: \( a_x(t) = \alpha t^2 \)
• In the y-direction: \( a_y(t) = \beta - \gamma t \)

These differences imply that the force acting on the rocket in each direction varies with time.

• For the x-direction, acceleration increases quadratically over time. This suggests a gradually increasing force impact as time passes.
• In the y-direction, due to \( \beta \) and \( \gamma \), the net acceleration is essentially reducing with time, due to a linearly decreasing factor \( \gamma t \).

Understanding these components helps in breaking the integral functions needed to find velocity.

Integration in Physics

In physics, integration is a fundamental calculus tool that helps determine how variables like velocity and position evolve over time. In our exercise, we integrated acceleration functions to find their corresponding velocities and subsequently, the positions.
Here's a simplified method to understand the process:

• For velocity: Integrate the acceleration functions from zero to a general time \( t \).
• Add the given initial values of velocity to these integrated functions, since motion begins with these initial velocities in each component.
• For position: Once velocity functions are determined, integrate them similarly to find how position changes with time.

The beauty of integration is it allows us to see these changes as continuous processes, which is essential when calculating projectile motion's trajectory like in this model rocket problem.

Velocity and Position Vectors

Velocity and position vectors describe the motion of an object in terms of speed and displacement, respectively. In the context of the exercise:

• Initial velocities given are \( v_{0x} = 1.00 \ m/s \) and \( v_{0y} = 7.00 \ m/s \).
• These vectors help us begin calculations for how position and velocity will evolve.

Through integration:

• Velocity vectors derived from the acceleration components show how the speed changes over time: \( v_x(t) \) and \( v_y(t) \).
• The position vectors show the trajectory or path: \( x(t) \) and \( y(t) \). These functions offer insights into both horizontal displacement and maximum height reached.

Understanding these vectors in terms of both magnitude and direction is key to solving kinematic problems effectively.