Problem 2 - College Physics

Question

At an air show, a jet plane has velocity components \(v_{x}=\) 625 \(\mathrm{km} / \mathrm{h}\) and \(v_{y}=415 \mathrm{km} / \mathrm{h}\) at time 3.85 \(\mathrm{s}\) and \(v_{x}=838 \mathrm{km} / \mathrm{h}\) and \(v_{y}=365 \mathrm{km} / \mathrm{h}\) at time 6.52 s. For this time interval, find(a) the \(x\) and \(y\) components of the plane's average acceleration and (b) the magnitude and direction of its average acceleration.

Step-by-Step Solution

Answer

The average acceleration components are 22.16 m/s² (x) and -5.20 m/s² (y), with a magnitude of 22.76 m/s² at -13.16°.

Step 1: Calculate Change in Velocity for X-Component

To find the average acceleration, we first calculate the change in velocity for the x-component. Given initial velocity \(v_{x1} = 625\, \text{km/h}\) and final velocity \(v_{x2} = 838\, \text{km/h}\), we have:\[\Delta v_x = v_{x2} - v_{x1} = 838 - 625 = 213 \, \text{km/h}.\]

Step 2: Convert Change in Velocity for X-Component to Meters per Second

Convert \(\Delta v_x\) into meters per second since the time is given in seconds. We use the conversion factor \(1\, \text{km/h} = \frac{1}{3.6}\, \text{m/s}\) to convert:\[\Delta v_x = 213 \times \frac{1}{3.6} = 59.17 \, \text{m/s}.\]

Step 3: Calculate Change in Velocity for Y-Component

Similarly, find the change in velocity for the y-component. Given initial velocity \(v_{y1} = 415\, \text{km/h}\) and final velocity \(v_{y2} = 365\, \text{km/h}\), we have:\[\Delta v_y = v_{y2} - v_{y1} = 365 - 415 = -50 \, \text{km/h}.\]

Step 4: Convert Change in Velocity for Y-Component to Meters per Second

Convert \(\Delta v_y\) to meters per second:\[\Delta v_y = -50 \times \frac{1}{3.6} = -13.89 \, \text{m/s}.\]

Step 5: Calculate the Time Interval

The time interval is the difference between the two given times:\[\Delta t = t_2 - t_1 = 6.52 - 3.85 = 2.67 \, \text{s}.\]

Step 6: Determine Average Acceleration for X-Component

The average acceleration for the x-component is given by the change in velocity divided by the time interval:\[a_{x, \text{avg}} = \frac{\Delta v_x}{\Delta t} = \frac{59.17}{2.67} = 22.16 \, \text{m/s}^2.\]

Step 7: Determine Average Acceleration for Y-Component

The average acceleration for the y-component is:\[a_{y, \text{avg}} = \frac{\Delta v_y}{\Delta t} = \frac{-13.89}{2.67} = -5.20 \, \text{m/s}^2.\]

Step 8: Calculate Magnitude of Average Acceleration

The magnitude of the average acceleration is found using the Pythagorean theorem:\[|a_{\text{avg}}| = \sqrt{(a_{x, \text{avg}})^2 + (a_{y, \text{avg}})^2} = \sqrt{(22.16)^2 + (-5.20)^2} = 22.76 \, \text{m/s}^2.\]

Step 9: Determine Direction of Average Acceleration

The direction \(\theta\) of the average acceleration relative to the x-axis is:\[\theta = \arctan\left(\frac{a_{y, \text{avg}}}{a_{x, \text{avg}}}\right) = \arctan\left(\frac{-5.20}{22.16}\right) = -13.16^\circ.\] This implies the direction is \(13.16^\circ\) below the positive x-axis.

Key Concepts

Velocity Components

When dealing with velocity, it's essential to consider its components, especially in two-dimensional motion. Velocity components are parts of the jet plane's velocity that act along the coordinate axes, such as the x-axis (horizontal) and y-axis (vertical). In this exercise, we observe that the plane's velocity at one instant

• along the x-axis is 625 km/h
• and along the y-axis is 415 km/h.

Later, the velocity components change to 838 km/h for the x-axis and 365 km/h for the y-axis. By breaking down the velocity into these components, we can analyze and calculate changes effectively. This approach allows for an easier computation of average acceleration and provides a clear understanding of how the velocity changes over time. Let's take a closer look at how these changes influence the average acceleration.

Change in Velocity

The change in velocity is a crucial concept when calculating acceleration. A change in velocity can occur in either component — the x-component or the y-component. To calculate the change in velocity, you simply subtract the initial velocity from the final velocity for each component.
Consider the x-component:

• Initial velocity: 625 km/h
• Final velocity: 838 km/h
• Change: 838 - 625 = 213 km/h

For the y-component:

• Initial velocity: 415 km/h
• Final velocity: 365 km/h
• Change: 365 - 415 = -50 km/h

These calculated changes in velocity allow us to proceed to the next step, which involves calculating acceleration. Remember to convert these changes into meters per second from kilometers per hour using the conversion factor to ensure consistency in units when calculating the acceleration.

Pythagorean Theorem

Once we have the average accelerations for both the x and y components, the Pythagorean theorem is our perfect tool to determine the resultant magnitude of the acceleration. This theorem states that, in a right triangle, the square of the hypotenuse (longest side) equals the sum of the squares of the other two sides.
In our scenario, think of:

• the x-component of acceleration as one side
• the y-component as the other side.

The formula is: \[|a_{\text{avg}}| = \sqrt{(a_{x, \text{avg}})^2 + (a_{y, \text{avg}})^2}\]So substitute the values of the components to find the magnitude of average acceleration. In practice, this means using hard numbers to calculate. For example, with given values: \(22.16 \, \text{m/s}^2\) and \(-5.20 \, \text{m/s}^2\), the magnitude calculated using Pythagorean theorem gives us a clearer picture of the overall effect - it's about figuring out the actual speed change considering both directions.

Direction of Acceleration

Knowing the direction of acceleration helps to understand how and where the velocity of an object is changing. To find the direction, we use the arctan function, which is part of trigonometry.
This involves calculating the direction angle \(\theta\) of the average acceleration relative to the x-axis. The formula used is: \[\theta = \arctan\left(\frac{a_{y, \text{avg}}}{a_{x, \text{avg}}}\right)\]This function helps determine the angle of movement considering the x and y components. Here, \(\theta\) indicates how much the path is altered from the horizontal axis. In this specific situation, \( -13.16^\circ \) means the acceleration vector has a small downward angle below the positive x-axis.
Keep this in mind: this angle measurement is crucial when translating mathematical results into real world directions, telling you precisely where the object's path is heading. By understanding the direction, predictions about future positions and velocities of moving objects become much more manageable.