Problem 40 - University Physics with Modern Physics

Question

A faulty model rocket moves in the \(xy\)-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 \(\alpha = 2.50 m/s^4, \beta = 9.00 m/s^2,\) and \(\gamma = 1.40 m/s^3\). At \(t = 0\) the rocket is at the origin and has velocity \(\vec{v}_0=v_0\hat{i} + v_{0y}\hat{j}\) with \(v_{0x}\) = 1.00 m/s and \(v_{0y}\) = 7.00 m/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

Answer

(a) Velocity: \(v_x(t) = 0.83t^3 + 1\), \(v_y(t) = 9t - 0.7t^2 + 7\); Position: \(x(t) = 0.208t^4 + t\), \(y(t) = 4.5t^2 - 0.233t^3 + 7t\). (b) Max height: calculate \(y(13.57)\). (c) Horizontal displacement: evaluate \(x(t)\) at return time found from the quadratic.

Step 1: Integrate Acceleration to Find Velocity Functions

To find the velocity as a function of time, we need to integrate the acceleration components with respect to time.\[a_x(t) = \alpha t^2 = 2.50 t^2 \v_x(t) = \int a_x(t) \, dt = \int 2.50 t^2 \, dt = \frac{2.50}{3} t^3 + C_x\]Given that \(v_{0x} = 1.00\, m/s\) at \(t=0\), we find \(C_x = 1.00\). Hence:\[v_x(t) = \frac{2.50}{3} t^3 + 1.00\]Similarly for the \(y\)-component:\[a_y(t) = \beta - \gamma t = 9.00 - 1.40t \v_y(t) = \int a_y(t) \, dt = \int (9.00 - 1.40t) \, dt = 9.00t - 0.70t^2 + C_y\]Given that \(v_{0y} = 7.00\, m/s\) at \(t=0\), we find \(C_y = 7.00\). Hence:\[v_y(t) = 9.00t - 0.70t^2 + 7.00\]

Step 2: Integrate Velocity Functions to Find Position Functions

To find the position as a function of time, integrate the velocity components with respect to time.\[x(t) = \int v_x(t) \, dt = \int \left(\frac{2.50}{3} t^3 + 1.00\right) \, dt = \frac{2.50}{12} t^4 + 1.00t + C_x'(t)\]At \(t=0\), \(x(0)=0\), hence \(C_x' = 0\). Thus:\[x(t) = \frac{2.50}{12} t^4 + 1.00t\] For the \(y\)-component:\[y(t) = \int v_y(t) \, dt = \int (9.00t - 0.70t^2 + 7.00) \, dt = 4.50t^2 - 0.233t^3 + 7.00t + C_y'(t)\]At \(t=0\), \(y(0)=0\), hence \(C_y' = 0\). Thus:\[y(t) = 4.50t^2 - 0.233t^3 + 7.00t\]

Step 3: Find Maximum Height

The maximum height is reached when the vertical velocity equals zero: \(v_y(t) = 0\).\[v_y(t) = 9.00t - 0.70t^2 + 7.00 = 0\]This is a quadratic equation can be solved for \(t\):\[0.70t^2 - 9.00t - 7.00 = 0\]Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=0.70\), \(b=-9.00\), and \(c=7.00\).\[t = \frac{9.00 \pm \sqrt{(-9.00)^2 - 4 \cdot 0.70 \cdot 7.00}}{2 \cdot 0.70}\]This gives \(t \approx 13.57\) seconds (we take the positive solution).Substitute \(t = 13.57\) into \(y(t)\) to find the maximum height:\[y(13.57) = 4.50(13.57)^2 - 0.233(13.57)^3 + 7.00(13.57)\]

Step 4: Calculate Horizontal Displacement when y=0

When the rocket returns to \(y = 0\), find \(t\) by solving:\[y(t) = 4.50t^2 - 0.233t^3 + 7.00t = 0\]Factor out the common term:\[t(4.50t - 0.233t^2 + 7.00) = 0\]The solutions will include \(t=0\) and a non-zero \(t\) when the rocket returns. Solve the quadratic equation:\[0.233t^2 - 4.50t - 7.00 = 0\]Using the quadratic formula with given values:\[t = \frac{4.50 \pm \sqrt{(-4.50)^2 - 4 \cdot 0.233 \cdot (-7.00)}}{2 \cdot 0.233}\]Select the positive \(t\) value which is not zero. Use this \(t\) in the expression for \(x(t)\) to find the horizontal displacement:\[x(t) = \frac{2.50}{12}(t^4) + 1.00t\] evaluated at the second positive \(t\).

Key Concepts

Kinematics

Kinematics is a crucial part of physics that deals with the description of motion, without considering the forces that cause it. In our exercise, we explore a faulty rocket in the xy-plane, analyzing how its position and velocity change over time. Important parameters to understand here are:

• Displacement: This refers to the change in position. It is a vector quantity, meaning it has both magnitude and direction.
• Velocity: This is the rate of change of displacement with respect to time. It's also vectorial, providing information on the speed and direction of the object's motion.
• Acceleration: Defined as the rate of change of velocity, it tells us how quickly an object speeds up or slows down. Acceleration can also have both horizontal and vertical components in the xy-plane, as seen in this exercise.

The mathematical model provided uses these elements to calculate the rocket's velocity and position over time, giving us a clearer picture of its motion through integration—a key method used in kinematics to derive one motion descriptor from another.

Integration in Physics

In physics, integration is a powerful tool, especially when dealing with varying rates of change, such as the scenario presented in kinematic equations. When we integrate acceleration, we find velocity; and when we integrate velocity, we obtain position. These integrations help us understand the complete trajectory of an object.
In our exercise, the rocket's acceleration components, represented by polynomials in terms of time, were integrated over time to obtain velocity functions. The integration of polynomial functions, like \(\alpha t^2\) for acceleration, follows a standard rule, where \(t^n\) is integrated to yield \((1/(n+1))t^{n+1}\). For example, \( a_x(t) = 2.50t^2\) becomes \( v_x(t) = \frac{2.50}{3}t^3 + C_x\) after integration. The constant \( C_x \) is determined using the provided initial conditions. This process illustrates how integration allows for the construction of dynamic motion models from known rates of change.

Quadratic Equations in Physics

Quadratic equations frequently appear in physics, especially in kinematics. They help solve for critical points where motion parameters like velocity or displacement change behavior.
In our rocket problem, we encounter quadratic equations when determining the maximum height. The rocket reaches its peak when its vertical velocity, \( v_y(t) \), equals zero. This equation becomes: \[ 0.70t^2 - 9.00t - 7.00 = 0 \] Such a quadratic equation is solved using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here 'a,' 'b,' and 'c' are coefficients from the quadratic equation. This provides time values at which key moments (like maximum height) occur. Picking the relevant positive time solution allows us to pin down precise motion characteristics.

Vector Components

Understanding vector components is crucial when dealing with motions in two or three dimensions. A vector is split into components along the coordinate axes, simplifying the analysis of motion. For the rocket example, both the acceleration and velocity are given in terms of their x and y vector components.
Breaking vectors such as acceleration into components helps solve physics problems, as it allows independent analysis of effects along each axis. For the x-component, \( a_x(t) \), the acceleration varies with \( t^2 \), primarily influencing the horizontal motion. Similarly, the y-component, \( a_y(t) \), impacts the vertical path of the rocket due to its linear time dependency.
Using component methods streamlines computation and conceptual understanding. Each component can be individually integrated and analyzed, producing linear equations that collectively form a vector description of motion. This separation clarifies how different forces and initial conditions impact the trajectory uniquely along each axis.