/*! 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 6 A dog running in an open field h... [FREE SOLUTION] | 91Ó°ÊÓ

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A dog running in an open field has components of velocity \(\boldsymbol{v}_{x}=2.6 \mathrm{m} / \mathrm{s}\) and \(v_{y}=-1.8 \mathrm{m} / \mathrm{s}\) at \(t_{1}=10.0 \mathrm{s}\) . For the time interval from \(t_{1}=10.0 \mathrm{s}\) to \(t_{2}=20.0 \mathrm{s}\) , the average acceleration of the dog has magnitude 0.45 \(\mathrm{m} / \mathrm{s}^{2}\) and direction \(31.0^{\circ}\) measured from the \(+x\) -axis toward the \(+y\) -axis. At \(t_{2}=20.0 \mathrm{s},\) (a) what are the \(x-\) and \(y\) -components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ?

Short Answer

Expert verified
(a) \( v_{x2} \approx 6.45 \) m/s, \( v_{y2} \approx 0.51 \) m/s. (b) \( v_2 \approx 6.47 \) m/s, \( 4.5^\circ \) from \(+x\). (c) Velocity increases and changes direction.

Step by step solution

01

Identify Initial Conditions

The initial conditions include the dog's velocity components at \( t_1 = 10.0 \,\mathrm{s} \): \( v_{x1} = 2.6 \,\mathrm{m/s} \) and \( v_{y1} = -1.8 \,\mathrm{m/s} \). The average acceleration has a magnitude of \( 0.45 \,\mathrm{m/s}^2 \) and a direction of \( 31.0^\circ \) from the \(+x\)-axis towards the \(+y\)-axis.
02

Resolve Acceleration into Components

The acceleration components are \( a_x = 0.45 \cos(31.0^\circ) \) and \( a_y = 0.45 \sin(31.0^\circ) \). Calculate these values to find \( a_x \approx 0.385 \) m/s² and \( a_y \approx 0.231 \) m/s².
03

Calculate Velocity Components at \( t_2 \)

Use the equations: \( v_{x2} = v_{x1} + a_x \Delta t \) and \( v_{y2} = v_{y1} + a_y \Delta t \), where \( \Delta t = t_2 - t_1 = 10 \) s. Calculate \( v_{x2} \approx 6.45 \) m/s and \( v_{y2} \approx 0.51 \) m/s.
04

Determine Magnitude of Final Velocity

The magnitude of the velocity at \( t_2 \) is \( \sqrt{v_{x2}^2 + v_{y2}^2} \). Substitute \( v_{x2} \) and \( v_{y2} \) to find the magnitude as \( v_{2} \approx 6.47 \) m/s.
05

Determine Direction of Final Velocity

The direction is given by \( \theta = \tan^{-1}(\frac{v_{y2}}{v_{x2}}) \). Compute \( \theta \approx \tan^{-1}(\frac{0.51}{6.45}) \), which results in \( \theta \approx 4.5^\circ \) from the \(+x\)-axis.
06

Sketch Velocity Vectors

Draw the initial velocity vector with components \( v_{x1} \) and \( v_{y1} \) and note that the vector points slightly downward due to the negative \( y \)-component. Draw the final velocity vector with components \( v_{x2} \) and \( v_{y2} \), pointing slightly upward, showing the change in direction due to the positive \( y \)-component.

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Key Concepts

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

Velocity Components
Understanding velocity components is crucial for analyzing motion in two dimensions. Whenever an object, like the dog in our problem, moves in a plane, we break down its velocity into horizontal and vertical parts called components. These components are expressed as \( v_x \) for the x-axis (horizontal) and \( v_y \) for the y-axis (vertical).
You can think of this as how fast the object moves along each axis separately.Initially, the dog has a velocity \( v_{x1} = 2.6 \text{ m/s} \) in the positive x-direction and \( v_{y1} = -1.8 \text{ m/s} \) in the negative y-direction. These components tell us that the dog is moving to the right while also slightly downward.
As time progresses and due to acceleration, these components change, resulting in new velocity components at a later time, \( t_2 = 20.0 \text{ s} \).
The final components are calculated using the formulas:
  • \( v_{x2} = v_{x1} + a_x \Delta t \)
  • \( v_{y2} = v_{y1} + a_y \Delta t \)
This approach helps us understand each dimension's contribution to the overall motion.
Average Acceleration
Average acceleration refers to the change in velocity over time. It describes how quickly an object's speed or direction changes. In the exercise, we have a magnitude of average acceleration given as \( 0.45 \text{ m/s}^2 \) with a specified direction of \( 31.0^\circ \) from the positive x-axis.
To work out the changes in velocity components, we need to resolve this acceleration into its x and y components using:
  • \( a_x = a \cos(\theta) \)
  • \( a_y = a \sin(\theta) \)
where \( a \) is the magnitude of acceleration and \( \theta \) is its direction angle.Using these calculations, we get \( a_x \approx 0.385 \text{ m/s²} \) and \( a_y \approx 0.231 \text{ m/s²} \). These components explain how much the velocity in the x and y directions changes, giving insight into the overall motion dynamics of the dog between the time intervals \( t_1 \) and \( t_2 \).
Vector Analysis
Vectors are essential tools in physics to describe quantities that have both magnitude and direction. In this problem, the velocity and acceleration are vectors. It's vital to treat each vector component as part of the larger whole, which represents the actual path or change occurring. To analyze these vectors, we often break them into components (as seen in velocity components). When dealing with changes in states, such as from one velocity to another, a vector analysis allows us to visually understand what's happening. This involves calculating:
  • Initial and Final Component Values
  • Magnitude and Direction Changes
  • Graphical Representation
Graphs or sketches of vectors help us visualize the initial and final velocities, showing the pathway and direction the dog takes at different instances. It emphasizes the shift from a slight downward movement to one in a more upward path as time progresses due to the acceleration applied. This transforms our basic numerical answer into an intuitive, clear representation of motion in two dimensions.
Trigonometry in Physics
Trigonometry in physics is an invaluable technique, especially for solving problems involving angles. In this problem, trigonometry aids in resolving vector magnitudes and directions into their components, using sine and cosine functions.
For the average acceleration given as \( 31.0^\circ \), we've used:
  • \( a_x = a \cos(\theta) \)
  • \( a_y = a \sin(\theta) \)
This allows for a complete description of motion in terms of its linear components along each axis. Beyond finding acceleration components, trigonometry also helps determine the overall direction of a resulting velocity, where:
  • \( \theta = \tan^{-1}(\frac{v_y}{v_x}) \)
By coupling trigonometry with vector analysis, we make solving two-dimensional motion problems achievable and clear, demonstrating how angles can translate a real-world scenario into mathematical terms.

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Most popular questions from this chapter

An airlane pilit wishes to fy due west. A wind of 80.0 \(\mathrm{km} / \mathrm{h}\) (about 50 \(\mathrm{mi} / \mathrm{h} )\) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 \(\mathrm{km} / \mathrm{h}\) (about 200 \(\mathrm{mi} / \mathrm{h} )\) , in which direction should the pilot head? (b) What is the speed of the plane over the ground? TIlustrate with a vector diagram.

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A projectile is launched with speed \(v_{0}\) at an angle \(\alpha_{0}\) above the horizontal. The launch point is a height \(h\) above the ground. (a) Show that if air resistance is ignored, the horizontal distance that the projectile travels before striking the ground is $$ x=\frac{v_{0} \cos \alpha_{0}}{g}\left(v_{0} \sin \alpha_{0}+\sqrt{v_{0}^{2} \sin ^{2} \alpha_{0}+2 g h}\right) $$ Verify that if the launch point is at ground level so that \(h=0\) , this is equal to the horizontal range \(R\) found in Example \(3.8 .\) (b) For the case where \(v_{0}=10 \mathrm{m} / \mathrm{s}\) and \(h=5.0 \mathrm{m},\) graph \(x\) as a function of launch angle \(\alpha_{0}\) for values of \(\alpha_{0}\) from \(0^{\circ}\) to \(90^{\circ} .\) Your graph should show that \(x\) is zero if \(\alpha_{0}=90^{\circ},\) but \(x\) is nonzero if \(\alpha_{0}=0\) ; explain why this is so. (c) We saw in Example 3.8 that for a projectile that lands at the same height from which it is launched, the horizontal range is maximum for \(\alpha_{0}=45^{\circ} .\) For the case graphed in part \((b),\) is the angle for maximum horizontal distance equal to, less than, or greater than \(45^{\circ} 7\) (This is a general result for the situation where a projectile is launched from a point higher than where it lands.)

A rock tied to a rope moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R \cos \omega t \quad y(t)=R \sin \omega t $$ where \(R\) and \(\omega\) are constants. (a) Show that the rock's distance from the origin is constant and equal to \(R-\) that is, that its path is a circle of radius \(R .(b)\) Show that at every point the rock's velocity is perpendicular to its position vector. (c) Show that the rock's acceleration is always opposite in direction to its position vector and has magnitude \(\omega^{2} R .\) (d) Show that the magnitude of the rock's velocity is constant and equal to \(\omega R\) . (e) Combine the results of parts \((\mathrm{c})\) and \((\mathrm{d})\) to show that the rock's acceleration has constant magnitude \(v^{2} / R .\)

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