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 \(\vec{\boldsymbol{v}}_{0}=v_{0 x} \hat{\boldsymbol{\imath}}+v_{0 y} \hat{\boldsymbol{J}}\) 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) Sketch the path of the rocket. (d) What is the horizontal displacement of the rocket when it returns to \(y=0 ?\)

Step by step solution

01

Integrating Acceleration to Find Velocity

To find the velocity components as functions of time, integrate the acceleration components with respect to time.The velocity in the x-direction is given by integrating \(a_x(t)\):\[ v_x(t) = \int a_x(t) \, dt = \int \alpha t^2 \, dt = \frac{\alpha t^3}{3} + C_1 \]At \(t=0\), \(v_x(0) = v_{0x}\), so \(C_1 = v_{0x}\). Substituting \(\alpha = 2.50\), \(v_{0x} = 1.00\):\[ v_x(t) = \frac{2.50 t^3}{3} + 1.00 \]The velocity in the y-direction is given by integrating \(a_y(t)\):\[ v_y(t) = \int a_y(t) \, dt = \int (\beta - \gamma t) \, dt = \beta t - \frac{\gamma t^2}{2} + C_2 \]At \(t=0\), \(v_y(0) = v_{0y}\), so \(C_2 = v_{0y}\). Substituting \(\beta = 9.00\), \(\gamma = 1.40\), \(v_{0y} = 7.00\):\[ v_y(t) = 9.00t - \frac{1.40 t^2}{2} + 7.00 \]

02

Integrating Velocity to Find Position

To find the position components as functions of time, integrate the velocity components with respect to time.The position in the x-direction is given by integrating \(v_x(t)\):\[ x(t) = \int v_x(t) \, dt = \int \left( \frac{2.50 t^3}{3} + 1.00 \right) \, dt = \frac{2.50 t^4}{12} + 1.00t + C_3 \]At \(t=0\), \(x(0) = 0\), so \(C_3 = 0\). Thus:\[ x(t) = \frac{2.50}{12} t^4 + 1.00t \]The position in the y-direction is given by integrating \(v_y(t)\):\[ y(t) = \int v_y(t) \, dt = \int \left( 9.00t - \frac{1.40 t^2}{2} + 7.00 \right) \, dt = \frac{9.00 t^2}{2} - \frac{1.40 t^3}{6} + 7.00t + C_4 \]At \(t=0\), \(y(0) = 0\), so \(C_4 = 0\). Thus:\[ y(t) = \frac{9.00 t^2}{2} - \frac{1.40 t^3}{6} + 7.00t \]

03

Calculating Maximum Height

To find the maximum height, set the vertical velocity \(v_y(t)\) to zero and solve for \(t\).\[ 0 = 9.00t - \frac{1.40 t^2}{2} + 7.00 \]Solving for \(t\) using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \text{ where } a = -\frac{1.40}{2}, b = 9.00, c = 7.00 \]\[ t = \frac{-9.00 \pm \sqrt{9.00^2 - 4 \times (-\frac{1.40}{2}) \times 7.00}}{-1.40} \]Calculate to find the time \(t_{max}\).Plug \(t_{max}\) back into \(y(t)\) to find the maximum height.

04

Calculate Horizontal Displacement at y = 0

The rocket returns to \(y = 0\) when its height function is zero again. Set \(y(t) = 0\) and solve for time \(t\) considering both the launch and return points:\[ 0 = \frac{9.00 t^2}{2} - \frac{1.40 t^3}{6} + 7.00t \]Factorize and solve for the positive \(t\) indicating the time of return.Once \(t\) is known, use it to find horizontal displacement \(x(t)\).

05

Sketch the Path

Plot calculated \(x(t)\) and \(y(t)\) values using a graphing tool or software to visualize the trajectory. This step forms a visual representation of the motion in the plane.

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Key Concepts

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

Acceleration

Acceleration is a fundamental concept in kinematics, describing how quickly an object's velocity changes with time. In this problem, we see that the acceleration of the rocket has components in both the x and y directions. Specifically:

• The x-component of acceleration is given by \(a_x(t) = \alpha t^2\), which means it increases with time as the square of time multiplied by the constant \(\alpha = 2.50 \text{ m/s}^4\).
• The y-component of acceleration is \(a_y(t) = \beta - \gamma t\), depicting a decrease in acceleration over time due to the \(-\gamma t\) term, where \(\beta = 9.00 \text{ m/s}^2\) and \(\gamma = 1.40 \text{ m/s}^3\).

In simple words, acceleration helps us understand how the rocket's speed changes as it moves along its path.

Velocity

Velocity tells us how fast the rocket is moving and in what direction. It is derived from integrating the acceleration components with respect to time. For this exercise, we found:

• The velocity in the x-direction is \(v_x(t) = \frac{2.50 t^3}{3} + 1.00\), where the initial velocity at time \(t=0\) is 1.00 m/s.
• The velocity in the y-direction is \(v_y(t) = 9.00t - \frac{1.40 t^2}{2} + 7.00\), starting at 7.00 m/s when \(t=0\).

Both of these equations show how the velocity changes over time and result from integrating the respective acceleration equations. Velocity is key to understanding the rocket's speed at any given moment during its flight.

Position

Position provides the location of the rocket in the x-y plane at a given time. It is derived by integrating the velocity components. For this problem:

• The position in the x-direction is \(x(t) = \frac{2.50}{12} t^4 + 1.00t\), demonstrating the path length covered horizontally.
• The position in the y-direction is \(y(t) = \frac{9.00 t^2}{2} - \frac{1.40 t^3}{6} + 7.00t\), showing the vertical distance covered.

These equations help us visualize and predict where the rocket will be at any time as it follows a specified trajectory.

Maximum Height

The maximum height attained by the rocket is a critical point, indicating the peak of its vertical motion. To find this point, we set the vertical velocity \(v_y(t) = 0\) and solve for time \(t\). This point gives the time when the rocket stops moving upward and starts descending.After calculating, this time is plugged back into the position function \(y(t)\) to determine the maximum height. At maximum height, the rocket momentarily has zero vertical velocity before gravity pulls it back down. This is essential for understanding the peak performance of the rocket's flight.

Trajectory

Trajectory refers to the path that the rocket follows through the air. By plotting the position functions \(x(t)\) and \(y(t)\), we can visualize this path on a graph. It shows how the rocket moves both horizontally and vertically with time.

• The shape of the trajectory is influenced by initial velocities, acceleration components, and forces like gravity.
• A thorough understanding of trajectory helps predict future positions and ensures safe and successful rocket travel.

Sketching the trajectory also aids in the visual learning process, making abstract concepts of motion more tangible and easier to grasp.