Problem 1 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

(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 1: 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:\[\Delta y = y_2 - y_1 = -0.5 \text{ m} - 3.4 \text{ m} = -3.9 \text{ m}\]

Step 2: 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:\[v_{y_{avg}} = \frac{\Delta y}{\Delta t} = \frac{-3.9 \text{ m}}{3.0 \text{ s}} = -1.3 \text{ m/s}\]

Step 3: Compute the Magnitude of Average Velocity

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

Step 4: 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.

Key Concepts

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.