/*! 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 116 A model airplane with mass 2.20 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A model airplane with mass 2.20 \(\mathrm{kg}\) moves in the \(x y\) -plane such that its \(x-\) and \(y\) -coordinates vary in time according to \(x(t)=\alpha-\beta t^{3}\) and \(y(t)=y t-\delta t^{2},\) where \(\alpha=1.50 \mathrm{m}, \beta=\) \(0.120 \mathrm{m} / \mathrm{s}^{3}, \gamma=3.00 \mathrm{m} / \mathrm{s},\) and \(\delta=1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) Calculate the \(x-\) and \(y\) -components of the net force on the plane as functions of time. (b) Sketch the trajectory of the airplane between \(t=0\) and \(t=3.00 \mathrm{s}\) , and draw on your sketch vectors showing the net force on the airplane at \(t=0, t=1.00 \mathrm{s}, t=2.00 \mathrm{s},\) and \(t=3.00 \mathrm{s}\) . For each of these times, relate the direction of the net force to the direction that the airplane is turning, and to whether the airplane is speeding up or slowing down (or neither). (c) What are the magnitude and direction of the net force at \(t=3.00 \mathrm{s} ?\)

Short Answer

Expert verified
(a) Forces: \( F_x(t) = -1.584t \), \( F_y(t) = -4.40 \). (b) Trajectory is parabolic; net force vectors show direction of turn, speeding/slowing effects. (c) At \( t=3.0 \, \text{s} \), force magnitude is \( 6.49 \, \text{N} \), direction \( 42.2^\circ \) down from x-axis.

Step by step solution

01

Find velocity components

First, determine the velocity components by differentiating the position functions with respect to time. The velocity in the x-direction, \( v_x(t) \), is the derivative of \( x(t) = \alpha - \beta t^3 \), giving:\[ v_x(t) = \frac{d}{dt}(\alpha - \beta t^3) = -3\beta t^2 \] Substitute \( \beta = 0.120 \, \text{m/s}^3 \):\[ v_x(t) = -3(0.120) t^2 = -0.36 t^2 \]The velocity in the y-direction, \( v_y(t) \), is the derivative of \( y(t) = \gamma t - \delta t^2 \), giving:\[ v_y(t) = \frac{d}{dt}(\gamma t - \delta t^2) = \gamma - 2\delta t \]Substitute \( \gamma = 3.00 \, \text{m/s} \) and \( \delta = 1.00 \, \text{m/s}^2 \):\[ v_y(t) = 3.00 - 2(1.00)t = 3.00 - 2t \]
02

Find acceleration components

Next, find the acceleration components by differentiating the velocity functions with respect to time. The acceleration in the x-direction, \( a_x(t) \), is the derivative of \( v_x(t) = -0.36 t^2 \), giving:\[ a_x(t) = \frac{d}{dt}(-0.36 t^2) = -0.72 t \]The acceleration in the y-direction, \( a_y(t) \), is the derivative of \( v_y(t) = 3.00 - 2t \), giving:\[ a_y(t) = \frac{d}{dt}(3.00 - 2t) = -2 \]
03

Calculate net force components

Use Newton's second law, \( F = m \, a \), to find the net force components. Given the mass \( m = 2.20 \, \text{kg} \), calculate the net force in the x-direction, \( F_x(t) \):\[ F_x(t) = m \, a_x(t) = 2.20(-0.72 t) = -1.584 t \]Similarly, calculate the net force in the y-direction, \( F_y(t) \):\[ F_y(t) = m \, a_y(t) = 2.20(-2) = -4.40 \]
04

Sketch the trajectory and analyze

Sketch the trajectory of the airplane using the position equations:- At \( t=0 \): \( x(0) = \alpha = 1.50 \) m, \( y(0) = 0 \)- At \( t=3.00 \): \( x(3) = 1.50 - 0.120 \times 3.00^3 = 1.50 - 3.24 = -1.74 \) m, \( y(3) = 3.00 \times 3 - 1.00 \times 3^2 = 9 - 9 = 0 \) mIdentify the positions at various time intervals and plot them. From the calculations, the trajectory resembles a parabolic path opening downwards.- At \( t=0 \), the force is entirely in the negative y-direction.- At \( t=1.00, 2.00, 3.00 \) s: draw vectors based on calculated forces. The direction indicates turning and speeding/slowing nature.
05

Calculate magnitude and direction of net force at t = 3 s

At \( t = 3.00 \, \text{s} \), substitute into net force equation components:\[ F_x(3) = -1.584 \times 3 = -4.752 \, \text{N} \]\[ F_y(3) = -4.40 \, \text{N} \]Calculate the magnitude of the total force \( F \):\[ F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-4.752)^2 + (-4.40)^2} \approx 6.49 \, \text{N} \]Find the direction using the angle \( \theta \) with respect to the x-axis:\[ \theta = \arctan\left(\frac{F_y}{F_x}\right) = \arctan\left(\frac{-4.40}{-4.752}\right) \approx 42.2^\circ \text{ (down from x-axis)} \]

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Key Concepts

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

Projectile Motion
Projectile motion describes the motion of an object or particle that is launched into the air and is subject to gravitational forces. Key factors influencing projectile motion include initial velocity, launch angle, and gravity. In this problem, the airplane's coordinates vary as functions of time, describing its trajectory through the air. The path it follows is parabolic due to the constant acceleration in the vertical direction, making it a classic example of projectile motion.

It's important to remember that projectile motion is a type of two-dimensional motion. The airplane's movement can be broken into two perpendicular components: horizontal (x-direction) and vertical (y-direction). These components are independent but occur simultaneously, each subject to their own set of forces and accelerations. Gravity mainly affects the vertical motion, while other forces can influence horizontal movement.
  • The horizontal component involves uniform motion since no acceleration acts horizontally.
  • The vertical component involves uniformly accelerated motion due to gravity.
Understanding projectile motion requires recognizing how these factors interplay to create the curved paths observed, like the parabolic trajectory calculated for the airplane here.
Acceleration
Acceleration describes the rate of change of velocity with respect to time. In this exercise, the acceleration components are derived by differentiating the velocity components of the airplane. It describes how quickly the airplane’s velocity changes both horizontally and vertically.

The horizontal acceleration, denoted as \( a_x(t) \), is calculated from the velocity function as \( a_x(t) = -0.72t \). This indicates that horizontal acceleration is not constant but changes linearly with time, leading to a gradual decrease in the plane's horizontal velocity.

On the other hand, the vertical acceleration, \( a_y(t) \), is constant at \( -2 \, \text{m/s}^2 \), reflecting a steady downward acceleration. This constant value is indicative of a uniformly accelerated motion found in the vertical component, typically a result of gravity, albeit adjusted for other influencing forces.
  • Horizontal acceleration: Linearly dependent on time, changing as the plane progresses.
  • Vertical acceleration: Constant, showing a steady influence of force.
Understanding how acceleration affects each component of motion is crucial in analyzing how the velocity and position evolve over time for objects like the airplane in projectile motion.
Force Components
In the realm of Newton's Laws of Motion, forces are vectors described by both magnitude and direction. The net force acting on an object is the vector sum of all individual forces acting on it. This exercise's task was to find these force components for the airplane, utilizing the mass and calculated accelerations of the plane.

The force components in both horizontal and vertical directions are determined using Newton's second law, \( F = ma \). For the airplane, these components are affected by its specific mass, \( m = 2.20 \, \text{kg} \), and the respective accelerations calculated.
  • Horizontal force, \( F_x(t) \), is time-dependent and calculated as \( F_x(t) = -1.584t \), demonstrating a decrease in horizontal force over time as the acceleration works against forward motion.
  • Vertical force, \( F_y(t) \), remains constant at \( -4.40 \, \text{N} \), consistent with the uniform, downward vertical acceleration.
Comprehending how these force components interact aids in predicting how the airplane moves, turns, and changes speed, especially in a situation where both gravity and other forces impact the trajectory.

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