Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A squirrel has \(x\)- and \(y\)-coordinates (1.1 m, 3.4 m) at time \(t_1\) = 0 and coordinates (5.3 m, -0.5 m) at time \(t_2\) = 3.0 s. For this time interval, find (a) the components of the average velocity, and (b) the magnitude and direction of the average velocity.

Short Answer

Expert verified
(a) \(v_{x_{avg}} = 1.4\) m/s, \(v_{y_{avg}} = -1.3\) m/s; (b) Magnitude: 1.91 m/s, Direction: -42.6°.

Step by step solution

01

Find the Components of Displacement

Determine the displacement components in the x- and y-directions. The displacement in the x-direction is the final x-coordinate minus the initial x-coordinate:\[\Delta x = x_2 - x_1 = 5.3 \text{ m} - 1.1 \text{ m} = 4.2 \text{ m}\]Similarly, the displacement in the y-direction is:\[\Delta y = y_2 - y_1 = -0.5 \text{ m} - 3.4 \text{ m} = -3.9 \text{ m}\]
02

Calculate the Components of Average Velocity

The average velocity components can be found by dividing the displacement components by the time interval. For the x-direction:\[v_{x_{avg}} = \frac{\Delta x}{\Delta t} = \frac{4.2 \text{ m}}{3.0 \text{ s}} = 1.4 \text{ m/s}\]And for the y-direction:\[v_{y_{avg}} = \frac{\Delta y}{\Delta t} = \frac{-3.9 \text{ m}}{3.0 \text{ s}} = -1.3 \text{ m/s}\]
03

Compute the Magnitude of Average Velocity

The magnitude of the average velocity is calculated using the Pythagorean theorem combining the x and y components:\[v_{avg} = \sqrt{v_{x_{avg}}^2 + v_{y_{avg}}^2} = \sqrt{(1.4 \text{ m/s})^2 + (-1.3 \text{ m/s})^2} = \sqrt{1.96 + 1.69} = \sqrt{3.65} \approx 1.91 \text{ m/s}\]
04

Determine the Direction of Average Velocity

The direction of the average velocity is given by the angle \(\theta\) with respect to the positive x-axis, using the arctangent function,\[\theta = \tan^{-1}\left(\frac{v_{y_{avg}}}{v_{x_{avg}}}\right) = \tan^{-1}\left(\frac{-1.3}{1.4}\right)\]Calculating this gives approximately \(\theta \approx -42.6^\circ\). The angle is measured clockwise from the positive x-axis.

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 Vaia!

Key Concepts

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

Displacement
Displacement refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. Here, displacement is calculated individually for each coordinate, and together they describe the overall change in position. To find the displacement for each component, you subtract the initial position from the final position:
  • For the x-direction: \[\Delta x = x_2 - x_1 = 5.3 \, \text{m} - 1.1 \, \text{m} = 4.2 \, \text{m}\]
  • For the y-direction: \[\Delta y = y_2 - y_1 = -0.5 \, \text{m} - 3.4 \, \text{m} = -3.9 \, \text{m}\]
The x-component of displacement indicates movement towards the right, while the negative y-component indicates movement downwards. The combination of these components gives the squirrel's overall displacement over the given time interval.
Velocity Components
Velocity components are crucial for understanding how an object's speed is distributed along different directions. For this problem, the velocity components represent how far the squirrel moves per second in the horizontal and vertical directions. They are found by dividing the displacement components by the time over which the motion occurs:
  • X-component of average velocity: \[v_{x_{avg}} = \frac{\Delta x}{\Delta t} = \frac{4.2 \, \text{m}}{3.0 \, \text{s}} = 1.4 \, \text{m/s}\]
  • Y-component of average velocity: \[v_{y_{avg}} = \frac{\Delta y}{\Delta t} = \frac{-3.9 \, \text{m}}{3.0 \, \text{s}} = -1.3 \, \text{m/s}\]
From these calculations, the squirrel's average horizontal velocity is 1.4 m/s to the right, and average vertical velocity is -1.3 m/s downward. This breakdown helps in visualizing how the total velocity is oriented in space.
Magnitude of Velocity
The magnitude of velocity gives an overall measure of speed, regardless of direction. It is a scalar quantity, calculated by combining the velocity components using the Pythagorean theorem. This gives the speed at which the squirrel is moving through space:\[v_{avg} = \sqrt{v_{x_{avg}}^2 + v_{y_{avg}}^2} = \sqrt{(1.4 \, \text{m/s})^2 + (-1.3 \, \text{m/s})^2}\]Calculating further, we have:\[v_{avg} = \sqrt{1.96 + 1.69} = \sqrt{3.65} \approx 1.91 \, \text{m/s}\]This value, 1.91 m/s, represents the actual speed of the squirrel over the time interval, regardless of the path it took from start to finish.
Direction of Velocity
The direction of velocity is expressed as an angle, giving the precise orientation of the velocity vector with respect to a reference line, usually the positive x-axis. It is calculated using the ratio of the velocity components:\[\theta = \tan^{-1}\left(\frac{v_{y_{avg}}}{v_{x_{avg}}}\right) = \tan^{-1}\left(\frac{-1.3}{1.4}\right)\]Solving for the angle gives:\[\theta \approx -42.6^\circ\]The negative angle indicates a direction clockwise from the positive x-axis. This helps visualize that the squirrel's movement was not straight but tilted downward as it moved horizontally.

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

Firemen use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of 25.0 m/s as it leaves the end of the hose and then exhibits projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 s to reach a building 45.0 m away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

The froghopper, \(Philaenus\) \(spumarius\), holds the world record for insect jumps. When leaping at an angle of 58.0\(^\circ\) above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (See \(Nature\), Vol. 424, July 31, 2003, p. 509.) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the earth's equator? Give your answer in m/s\(^2\) and as a fraction of \(g\). (b) If \(a_{rad}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north-south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (\(Note\): Even on cloudy nights, many birds can navigate by using the earth's magnetic field to fix the north-south direction.)

A sly 1.5-kg monkey and a jungle veterinarian with a blow-gun loaded with a tranquilizer dart are 25 m above the ground in trees 70 m apart. Just as the veterinarian shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. What must the minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground?

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.

Sign-up for free