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Chapter 3: Problem 1

A squirrel has \(x\) - and \(y\) -coordinates \((1.1 \mathrm{m}, 3.4 \mathrm{m})\) at time \(t_{1}=0\) and coordinates \((5.3 \mathrm{m},-0.5 \mathrm{m})\) at time \(t_{2}=3.0 \mathrm{s}\) . 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
The average velocity components are \(1.4 \mathrm{m/s}\) and \(-1.3 \mathrm{m/s}\). The magnitude is \(1.91 \mathrm{m/s}\) with a direction of \(-42.0^\circ\).

Step by step solution

01

Determine change in x and y coordinates

The change in the x-coordinate, denoted as \[\Delta x = x_2 - x_1\]Where \(x_1 = 1.1 \mathrm{m}\) and \(x_2 = 5.3 \mathrm{m}\). Calculating this gives \[\Delta x = 5.3 \mathrm{m} - 1.1 \mathrm{m} = 4.2 \mathrm{m}\].Similarly, the change in the y-coordinate is\[\Delta y = y_2 - y_1\]Where \(y_1 = 3.4 \mathrm{m}\) and \(y_2 = -0.5 \mathrm{m}\). Calculating this gives \[\Delta y = -0.5 \mathrm{m} - 3.4 \mathrm{m} = -3.9 \mathrm{m}\].
02

Calculate average velocity components

The average velocity components, \(v_{x, avg}\) and \(v_{y, avg}\), are given by \[v_{x, avg} = \frac{\Delta x}{\Delta t} = \frac{4.2 \mathrm{m}}{3.0 \mathrm{s}} = 1.4 \mathrm{m/s}\]and\[v_{y, avg} = \frac{\Delta y}{\Delta t} = \frac{-3.9 \mathrm{m}}{3.0 \mathrm{s}} = -1.3 \mathrm{m/s}\].
03

Calculate magnitude of average velocity

The magnitude of the average velocity \(v_{avg}\) is given by \[v_{avg} = \sqrt{(v_{x, avg})^2 + (v_{y, avg})^2} = \sqrt{(1.4)^2 + (-1.3)^2}\]Calculating this gives\[v_{avg} = \sqrt{1.96 + 1.69} = \sqrt{3.65} \approx 1.91 \mathrm{m/s}\].
04

Calculate direction of average velocity

The direction \(\theta\) of the average velocity relative to the positive x-axis can be found using \[\theta = \arctan\left(\frac{v_{y, avg}}{v_{x, avg}}\right) = \arctan\left(\frac{-1.3}{1.4}\right)\]Calculating this gives\[\theta \approx -42.0^\circ\].The negative angle indicates that the direction is below the positive x-axis.

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Key Concepts

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

Understanding Coordinates
Coordinates are a way to represent the position of a point in a two-dimensional space. In this exercise, the squirrel's position at different times is given as pairs of coordinates, \(x, y\). These coordinates indicate its position along the horizontal (x-axis) and vertical (y-axis) directions. Think of the x-coordinate as how far left or right the squirrel is from a starting point, and the y-coordinate as how far up or down it is.
  • The initial position is \( (1.1 \, \text{m}, 3.4 \, \text{m}) \).
  • The final position is \( (5.3 \, \text{m}, -0.5 \, \text{m}) \).
These coordinates allow us to calculate other important aspects of motion, like the change in position and the average velocity.
Calculating Change in Position
Change in position is a crucial concept when studying motion. It tells us how far the squirrel has moved in either direction. We calculate this by finding the difference between the final and initial coordinates.
  • The change in x-coordinate, \( \Delta x = x_2 - x_1 = 5.3 \, \text{m} - 1.1 \, \text{m} = 4.2 \, \text{m} \).
  • The change in y-coordinate, \( \Delta y = y_2 - y_1 = -0.5 \, \text{m} - 3.4 \, \text{m} = -3.9 \, \text{m} \).
These differences (\(\Delta x\)and \(\Delta y\)) reveal the actual movement the squirrel made along each axis. Understanding these changes is the first step towards calculating the average velocity.
Magnitude and Direction of Average Velocity
The magnitude and direction of the average velocity provide complete information about the squirrel's movement over time. The magnitude refers to how fast the squirrel is moving overall, while the direction indicates the path it's taking.
First, we calculate the magnitude using the formula:\[v_{avg} = \sqrt{(v_{x, avg})^2 + (v_{y, avg})^2} = \sqrt{(1.4)^2 + (-1.3)^2} = \sqrt{3.65} \approx 1.91 \, \text{m/s} \\]This value represents the squirrel's average speed. For direction, we determine the angle using:\[\theta = \arctan\left(\frac{v_{y, avg}}{v_{x, avg}}\right) = \arctan\left(\frac{-1.3}{1.4}\right) \approx -42.0^\circ \\]This angle is measured from the positive x-axis. A negative sign means the direction is below this axis. Together, these calculations offer a complete snapshot of the squirrel's motion.

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Most popular questions from this chapter

A 124 -kg balloon carrying a 22 -kg basket is descending with a constant downward velocity of 20.0 \(\mathrm{m} / \mathrm{s} . \mathrm{A} 1.0\) -kg stone is thrown from the basket with an initial velocity of 15.0 \(\mathrm{m} / \mathrm{s}\) perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.00 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 \(\mathrm{m} / \mathrm{s} .\) (a) How high was the balloon when the rock was thrown out? (b) How high is the balloon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.

A water hose is used to fill a large cylindrical storage tank of diameter \(D\) and height 2\(D .\) The hose shoots the water at \(45^{\circ}\) the horizontal from the same level as the base of the tank and is a distance 6\(D\) away (Fig. P3.60). For what range of launch speeds \(\left(v_{0}\right)\) will the water enter the tank? Ignore air resistance, and express your answer in terms of \(D\) and \(g .\)

A daring 510 -N swimmer dives off a cliff with a running horizontal leap, as shown in Fig. E3.10. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 \(\mathrm{m}\) wide and 9.00 \(\mathrm{m}\) below the top of the cliff?

A Ferris wheel with radius 14.0 \(\mathrm{m}\) is turning about a horizontal axis through its center (Fig. E3.29). The linear speed of a passenger on the rim is constant and equal to 7.00 \(\mathrm{m} / \mathrm{s} .\) What are the magnitude and direction of the passenger's acceleration as she passes through (a) the lowest point in her circular motion? (b) The highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 10.0 \(\mathrm{ft}\) above the ground, and the ball is kicked from ground level, 36.0 \(\mathrm{ft}\) horizontally from the bar (Fig. P3.62). Football regulations are stated in English units, but convert them to SI units for this problem.(a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it is to just clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) .
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