Problem 1 - College Physics

Question

A meteor streaking through the night sky is located with radar. At point \(A\) its coordinates are \((5.00 \mathrm{km}, 1.20 \mathrm{km}),\) and 1.14 s later it has moved to point \(B\) with coordinates \((6.24 \mathrm{km},\)0.925 \(\mathrm{km} ) .\) Find (a) the \(x\) and \(y\) components of its average velocity between \(A\) and \(B\) and (b) the magnitude and direction of its average velocity between these two points.

Step-by-Step Solution

Answer

The average velocity components are \(v_x = 1.088\, \text{km/s}\) and \(v_y = -0.241\, \text{km/s}\); magnitude is \(1.114\, \text{km/s}\) at \(-12.5^\circ\) below the x-axis.

Step 1: Calculate the change in x-coordinate

To find the change in the x-coordinate (\(\Delta x\)), subtract the initial x-coordinate at point A from the x-coordinate at point B:\(\Delta x = x_B - x_A = 6.24\, \text{km} - 5.00\, \text{km} = 1.24\, \text{km}\).This is the component of the change in position along the x-axis.

Step 2: Calculate the change in y-coordinate

To find the change in the y-coordinate (\(\Delta y\)), subtract the initial y-coordinate at point A from the y-coordinate at point B:\(\Delta y = y_B - y_A = 0.925\, \text{km} - 1.20\, \text{km} = -0.275\, \text{km}\).This is the component of the change in position along the y-axis.

Step 3: Compute average velocity components

Using the changes in coordinates from steps 1 and 2, find the x and y components of the average velocity:- The x-component of average velocity is \(v_{x, \text{avg}} = \frac{\Delta x}{\Delta t} = \frac{1.24\, \text{km}}{1.14\, \text{s}} \approx 1.088\, \text{km/s}\).- The y-component of average velocity is \(v_{y, \text{avg}} = \frac{\Delta y}{\Delta t} = \frac{-0.275\, \text{km}}{1.14\, \text{s}} \approx -0.241\, \text{km/s}\).

Step 4: Calculate the magnitude of the average velocity

The magnitude of the average velocity vector is calculated using the Pythagorean theorem:\(v_{\text{avg}} = \sqrt{(v_{x, \text{avg}})^2 + (v_{y, \text{avg}})^2} = \sqrt{(1.088)^2 + (-0.241)^2}\, \text{km/s}\).\(\approx \sqrt{1.184 + 0.058081}\, \text{km/s} = \sqrt{1.242081}\, \text{km/s} \approx 1.114\, \text{km/s}\).

Step 5: Determine the direction of the average velocity

The direction of the average velocity is given by the angle \(\theta\) above the positive x-direction, calculated using:\(\theta = \arctan\left(\frac{v_{y, \text{avg}}}{v_{x, \text{avg}}}\right) = \arctan\left(\frac{-0.241}{1.088}\right)\).\(\approx \arctan(-0.221)\).Using inverse tangent, \(\theta \approx -12.5^\circ\), which means the direction is approximately 12.5 degrees below the positive x-axis.

Key Concepts

Components of Velocity

When dealing with the motion of objects, such as a meteor streaking across the sky, understanding the components of velocity helps in visualizing how the object moves in different directions.
To find the components of velocity, we need to look at a change in position over time along each axis:

• **x-component**: This is the change along the horizontal axis. To find it, you calculate the difference between the x-coordinates of the two points. In this case: \[\Delta x = x_B - x_A = 6.24 \text{ km} - 5.00 \text{ km} = 1.24 \text{ km}.\]
• **y-component**: The change along the vertical axis. Similarly, derive this by subtracting the y-coordinates: \[\Delta y = y_B - y_A = 0.925 \text{ km} - 1.20 \text{ km} = -0.275 \text{ km}.\]

Next, divide each of these changes by the time duration (1.14 seconds) to obtain the average velocity components:

• x-component of average velocity: \[v_{x, \text{avg}} = \frac{1.24 \text{ km}}{1.14 \text{ s}} \approx 1.088 \text{ km/s}.\]
• y-component of average velocity:\[v_{y, \text{avg}} = \frac{-0.275 \text{ km}}{1.14 \text{ s}} \approx -0.241 \text{ km/s}.\]

This breakdown into components helps us understand how the meteor traverses horizontally and vertically through the sky.

Magnitude of Velocity

To get a sense of how fast the meteor is actually moving, we need to calculate the magnitude of its velocity. This involves finding the 'length' of the velocity vector using the Pythagorean theorem.
Think of this as the hypotenuse in a right triangle, where each leg corresponds to one of the velocity components we computed earlier.

• The formula to determine the magnitude is:\[v_{\text{avg}} = \sqrt{(v_{x, \text{avg}})^2 + (v_{y, \text{avg}})^2}.\]

Plug in the values we found:\[v_{\text{avg}} = \sqrt{(1.088 \text{ km/s})^2 + (-0.241 \text{ km/s})^2} \approx \sqrt{1.184 + 0.058081}\]which simplifies to:\[v_{\text{avg}} \approx \sqrt{1.242081} \approx 1.114 \text{ km/s}.\]This tells us that despite moving in both the x and y directions, the meteor has an average speed of about 1.114 km/s. The magnitude doesn't tell us the direction, only how fast it travels.

Direction of Velocity

Knowing how fast an object is moving is valuable, but understanding the direction is essential to fully describe its motion. The direction of velocity indicates the angle at which the object moves relative to the horizontal axis.
To find this direction, use the tangent of the angle, which relates the perpendicular velocity components:

• The direction, \(\theta\), is calculated using:\[\theta = \arctan \left(\frac{v_{y, \text{avg}}}{v_{x, \text{avg}}}\right).\]

Substitute in the values for the velocity components:\[\theta = \arctan \left(\frac{-0.241}{1.088}\right) \approx \arctan (-0.221) \approx -12.5^\circ.\]This negative angle indicates the direction is below the positive x-axis, meaning the meteor is moving 12.5 degrees downward relative to the horizontal.
Understanding this angle helps determine where the meteor is headed from its initial point of detection.