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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 \((\mathrm{b})\) the magnitude and direction of the average velocity.

Short Answer

Expert verified
The average velocity components are \(1.4 \text{ m/s}\) (x-dir) and \(-1.3 \text{ m/s}\) (y-dir); magnitude is \(1.91 \text{ m/s}\) with a direction of \(-43.6^\circ\).

Step by step solution

01

Understand the Given Information

The squirrel's initial coordinates are \((1.1 \text{ m}, 3.4 \text{ m})\) and final coordinates are \((5.3 \text{ m}, -0.5 \text{ m})\). The initial time is \(t_1 = 0\) seconds and the final time is \(t_2 = 3.0\) seconds.
02

Calculate the Change in x and y Coordinates

Calculate the differences in the x and y coordinates. The change in the x-coordinate is \(\Delta x = x_2 - x_1 = 5.3 \text{ m} - 1.1 \text{ m} = 4.2 \text{ m}\). The change in the y-coordinate is \(\Delta y = y_2 - y_1 = -0.5 \text{ m} - 3.4 \text{ m} = -3.9 \text{ m}\).
03

Calculate the Components of the Average Velocity

The average velocity component in the x-direction is \(v_{x, \text{avg}} = \frac{\Delta x}{\Delta t} = \frac{4.2 \text{ m}}{3.0 \text{ s}} = 1.4 \text{ m/s}\). The average velocity component in the y-direction is \(v_{y, \text{avg}} = \frac{\Delta y}{\Delta t} = \frac{-3.9 \text{ m}}{3.0 \text{ s}} = -1.3 \text{ m/s}\).
04

Calculate the Magnitude of the Average Velocity

The magnitude of the average velocity is given by \(v_{\text{avg}} = \sqrt{(v_{x, \text{avg}})^2 + (v_{y, \text{avg}})^2}\). Substituting the values, we get \(v_{\text{avg}} = \sqrt{(1.4)^2 + (-1.3)^2} = \sqrt{1.96 + 1.69} = \sqrt{3.65} \approx 1.91 \text{ m/s}\).
05

Calculate the Direction of the Average Velocity

The direction (angle \(\theta\)) of the average velocity with respect to the positive x-axis is given by \(\theta = \tan^{-1} \left(\frac{v_{y, \text{avg}}}{v_{x, \text{avg}}}\right)\). Substituting the values, \(\theta = \tan^{-1} \left(\frac{-1.3}{1.4}\right)\). Calculate using a calculator: \(\theta \approx \tan^{-1} (-0.929) \approx -43.6^\circ\). The negative angle indicates 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.

Velocity Components
When dealing with problems involving motion, it's essential to break down any movement into its fundamental parts, or components. In two dimensions, these components are typically along the x-axis (horizontal) and y-axis (vertical). This breakdown helps us better analyze how an object, like a squirrel in our exercise, moves in each direction independently.
To find the average velocity components, you calculate both the individual changes in the x and y directions over a specified time period:
  • \( v_{x, \text{avg}} = \frac{\Delta x}{\Delta t} \)
  • \( v_{y, \text{avg}} = \frac{\Delta y}{\Delta t} \)
By computing these elements, we determine how fast the squirrel is moving horizontally and vertically between the starting and ending coordinates. This calculates the rate of change of position, or velocity, in their respective directions.
Coordinate Geometry
Coordinate geometry allows us to precisely describe the position and trajectory of objects in space. By using a simple (x, y) coordinate system, we can track movement and calculate distances and direction.
For our exercise, we first note the starting point \((1.1 \text{ m}, 3.4 \text{ m})\) and ending point \((5.3 \text{ m}, -0.5 \text{ m})\) of the squirrel's path. Within this framework, the change in position or displacement is readily seen:
  • The change in the x-coordinate (horizontal distance) is \(\Delta x\).
  • The change in the y-coordinate (vertical distance) is \(\Delta y\).
By calculating these displacements, we easily derive path-related details, such as the velocity components, using these coordinates as a basis. Finally, coordinate geometry emphasizes the relationship between the numerical values and the graphical position, enhancing our understanding of the motion.
Magnitude and Direction of Velocity
Velocity isn't just about how fast something moves; it defines both speed and direction. To fully understand the squirrel's velocity, we must determine its magnitude and direction.Magnitude of Velocity
The magnitude of velocity gives us the overall speed irrespective of direction. Using the velocity components, the magnitude can be calculated through the Pythagorean theorem:
  • \[ v_{\text{avg}} = \sqrt{(v_{x, \text{avg}})^2 + (v_{y, \text{avg}})^2} \]
In this example, the computation results in approximately 1.91 m/s, showing the pace at which the squirrel moves along its trajectory.Direction of Velocity
The direction angle \(\theta\) indicates which way the object is heading relative to the positive x-axis. Calculated through the inverse tangent function, the direction provides context to the speed:
  • \( \theta = \tan^{-1} \left(\frac{v_{y, \text{avg}}}{v_{x, \text{avg}}}\right) \)
This exercise yields \(-43.6^\circ\), which means the direction is diagonally southward, or below the positive x-axis. Together, the magnitude and direction fully describe the velocity vector, painting a clear picture of the squirrel's movement.

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