/*! 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

91Ó°ÊÓ

A dog running in an open field has components of velocity \(v_{x}=2.6 \mathrm{m} / \mathrm{s}\) and \(v_{y}=-1.8 \mathrm{m} / \mathrm{s}\) at \(t_{1}=10.0 \mathrm{s}\) . For the time interval from \(t_{1}=10.0\) s to \(t_{2}=20.0\) 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
At \(t_2\), velocity components are 6.46 m/s (x) and 0.51 m/s (y); magnitude is 6.48 m/s and direction is 4.5° above the x-axis.

Step by step solution

01

Determine Acceleration in Components

Extract the acceleration components (\(a_x\) and \(a_y\)) using the given magnitude and direction. Use trigonometric functions to find:\[a_x = a \cos(31^\circ) = 0.45 \cos(31^\circ) = 0.386 \ m/s^2\]\[a_y = a \sin(31^\circ) = 0.45 \sin(31^\circ) = 0.231 \ m/s^2\]
02

Calculate Change in Velocity Components

With acceleration components calculated, use the relation of acceleration to change in velocity, where:\[\Delta v_x = a_x \Delta t\]\[\Delta v_y = a_y \Delta t\]Here, \(\Delta t = t_2 - t_1 = 20s - 10s = 10s\):\[\Delta v_x = 0.386 \times 10 = 3.86 \ m/s\]\[\Delta v_y = 0.231 \times 10 = 2.31 \ m/s\]
03

Calculate Velocity Components at t2

Use the initial velocities at \(t_1\) and add the changes to get the velocities at \(t_2\):\[v_{x,t2} = v_{x,t1} + \Delta v_x = 2.6 + 3.86 = 6.46 \ m/s\]\[v_{y,t2} = v_{y,t1} + \Delta v_y = -1.8 + 2.31 = 0.51 \ m/s\]
04

Calculate Resultant Velocity Magnitude at t2

Find the magnitude of the resultant velocity at \(t_2\):\[v_{t2} = \sqrt{v_{x,t2}^2 + v_{y,t2}^2} = \sqrt{6.46^2 + 0.51^2} = 6.48 \ m/s\]
05

Determine Direction of Velocity at t2

Calculate the direction of the velocity vector using the inverse tangent function:\[\theta_{t2} = \tan^{-1}\left(\frac{v_{y,t2}}{v_{x,t2}}\right) = \tan^{-1}\left(\frac{0.51}{6.46}\right) \approx 4.5^\circ\]
06

Describe and Compare Vectors at t1 and t2

- At \(t_1\), the velocity vector points with components \(v_x = 2.6 \ m/s\) and \(v_y = -1.8 \ m/s\).- At \(t_2\), the velocity vector points with components \(v_x = 6.46 \ m/s\) and \(v_y = 0.51 \ m/s\), indicating a change.- The vector has increased in magnitude to 6.48 m/s and turned towards the x-axis due to upward acceleration.

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.

Average Acceleration
Average acceleration is a measure of how much the velocity of an object changes over a period of time. It is expressed as a vector, which means it has both magnitude and direction. The formula to find average acceleration is given by:
  • \( \mathbf{a}_{avg} = \frac{\Delta \mathbf{v}}{\Delta t} \)
where \( \Delta \mathbf{v} \) is the change in velocity and \( \Delta t \) is the time interval.
In the given exercise, the dog experiences an average acceleration with a magnitude of 0.45 \( \text{m/s}^2 \) and a specific direction of 31 degrees. To break this down into comprehensible pieces, it is important to understand that acceleration can be dissected into its components using trigonometry.
The horizontal (\(a_x\)) and vertical (\(a_y\)) components of acceleration can be found by:
  • \( a_x = a \cos(31^\circ) \)
  • \( a_y = a \sin(31^\circ) \)
These allow us to calculate how much each individual component of velocity is affected over the time interval.
Velocity Components
Understanding velocity components is essential for solving problems that involve motion in two dimensions, such as the movement of the dog in the open field. Velocity is also a vector, which means it has both magnitude and direction. It can be separated into two perpendicular components – one along the x-axis and one along the y-axis.
Initially, the dog's velocity components are given as \(v_{x}=2.6 \ \text{m/s}\) and \(v_{y}=-1.8 \ \text{m/s}\). When the dog experiences acceleration, these components change over time. The change in the velocity components from \(t_1\) to \(t_2\) can be calculated using the acceleration components and the duration of time over which they act:
  • \( \Delta v_x = a_x \Delta t \)
  • \( \Delta v_y = a_y \Delta t \)
Here, \( \Delta t = 10 \ \text{s} \). Knowing the initial velocities and the changes, we can find the new velocities at \(t_2\):
  • \( v_{x,t2} = v_{x,t1} + \Delta v_x \)
  • \( v_{y,t2} = v_{y,t1} + \Delta v_y \)
In this example, the velocity in the y-direction increases, illustrating how the dog’s upward acceleration affects its motion.
Vector Analysis
Vector analysis is a powerful tool in physics that allows us to mathematically describe how objects in motion move and change direction. Vectors not only have magnitude, but they also have a direction, which is crucial for fully understanding two-dimensional motion.
To analyze vectors, such as velocity vectors, you can visualize them as arrows in a coordinate system. The length of the arrow represents the magnitude, while the direction of the arrow indicates the vector's direction. In this problem, the velocity vectors at \(t_1\) and \(t_2\) highlight how the dog’s movement evolves over time.
Vectors can be added and subtracted not by simple arithmetic, but by looking at their components along each axis. For the dog’s velocity at \(t_2\), once we have both component values \(v_{x,t2}\) and \(v_{y,t2}\), we can compute the overall speed, or resultant velocity, using the Pythagorean theorem:
  • \( v_{t2} = \sqrt{v_{x,t2}^2 + v_{y,t2}^2} \)
Moreover, the direction of this new vector is found using:
  • \( \theta_{t2} = \tan^{-1}\left(\frac{v_{y,t2}}{v_{x,t2}}\right) \)
This approach shows how motion can be broken down and comprehensively understood using vector analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An elevator is moving upward at a constant speed of2.50 \(\mathrm{m} / \mathrm{s} .\) A bolt in the elevator ceiling 3.00 \(\mathrm{m}\) above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings of the building? (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. P3.67). The takeoff ramp was inclined at \(53.0^{\circ},\) the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

A rock is thrown from the roof of a building with a velocity \(v_{0}\) at an angle of \(\alpha_{0}\) from the horizontal. The building has height \(h .\) You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground, and show that this speed is independent of \(\alpha_{0}\) .

At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravity") on test pilots and astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and the astronaut is strapped in at the other end. Suppose that he is aligned along the arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this machine is typically 12.5\(g\) . (a) How fast must the astronaut's head be moving to experience this maximum acceleration? (b) What is the difference between the acceleration of his head and feet if the astronaut is 2.00 \(\mathrm{m}\) tall? (c) How fast in rpm (rev/min) is the arm turning to produce the maximum sustained acceleration?

A rocket is fired at an angle from the top of a tower of height \(h_{0}=50.0 \mathrm{m} .\) Because of the design of the engines, its position coordinates are of the form \(x(t)=A+B t^{2}\) and \(y(t)=C+D t^{3},\) where \(A, B, C,\) and \(D\) are constants. Furthermore, the acceleration of the rocket 1.00 s after firing is \(\vec{a}=(4.00 \hat{\imath}+3.00 \hat{J}) \mathrm{m} / \mathrm{s}^{2} .\) Take the origin of coordinates to be at the base of the tower. (a) Find the constants \(A, B, C,\) and \(D,\) including their SI units. (b) At the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) What are the \(x\) -and \(y\) -components of the rocket's velocity 10.0 s after it is fired, and how fast is it moving? (d) What is the position vector of the rocket 10.0 s after it is fired?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.