Question

A jet plane is flying at a constant altitude. At time \(t_1\) = 0, it has components of velocity \(v_x\) = 90 m/s, \(v_y\) = 110 m/s. At time \(t_2\) = 30.0 s, the components are \(v_x\) = -170 m/s, \(v_y\) = 40 m/s. (a) Sketch the velocity vectors at \(t_1\) and \(t_2\). How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

Short answer

The average acceleration's magnitude is 8.98 m/s² at an angle of 15.1° below the positive x-axis.

Step-by-step solution

Identify the Components of Velocity

At time \( t_1 = 0 \), the components of velocities are given as \( v_{x1} = 90 \) m/s and \( v_{y1} = 110 \) m/s. At time \( t_2 = 30 \) s, these components become \( v_{x2} = -170 \) m/s and \( v_{y2} = 40 \) m/s.

Calculate the Change in Velocity Components

Calculate the change in the velocity components from \( t_1 \) to \( t_2 \). Change in \( v_x \) is \( \Delta v_x = v_{x2} - v_{x1} = -170 \, \text{m/s} - 90 \, \text{m/s} = -260 \, \text{m/s} \). Change in \( v_y \) is \( \Delta v_y = v_{y2} - v_{y1} = 40 \, \text{m/s} - 110 \, \text{m/s} = -70 \, \text{m/s} \).

Calculate the Components of Average Acceleration

The average acceleration is given by the change in velocity divided by the change in time. So, the average acceleration in the x-direction is \( a_{x, \text{avg}} = \frac{\Delta v_x}{\Delta t} = \frac{-260 \, \text{m/s}}{30.0 \, \text{s}} = -8.67 \, \text{m/s}^2 \). The average acceleration in the y-direction is \( a_{y, \text{avg}} = \frac{\Delta v_y}{\Delta t} = \frac{-70 \, \text{m/s}}{30.0 \, \text{s}} = -2.33 \, \text{m/s}^2 \).

Calculate the Magnitude of the Average Acceleration

The magnitude of the average acceleration can be found using the Pythagorean theorem: \( a_{\text{avg}} = \sqrt{a_{x, \text{avg}}^2 + a_{y, \text{avg}}^2} = \sqrt{(-8.67)^2 + (-2.33)^2} \, \text{m/s}^2 \). Therefore, \( a_{\text{avg}} = \sqrt{75.17 + 5.43} \approx 8.98 \, \text{m/s}^2 \).

Determine the Direction of the Average Acceleration

The direction \( \theta \) of the average acceleration with respect to the x-axis can be found using the tangent function: \( \theta = \arctan \left( \frac{a_{y, \text{avg}}}{a_{x, \text{avg}}} \right) = \arctan \left( \frac{-2.33}{-8.67} \right) \). This calculates to \( \theta = 15.1^\circ \) above the negative x-axis (or equivalently, below the positive x-axis).

Sketch and Compare the Velocity Vectors

For the sketch, draw the velocity vectors with appropriate directions: \( v_1 \) at \( t_1 \) pointing in the positive quadrant, and \( v_2 \) at \( t_2 \) pointing in the negative x-direction with a smaller y-component. They differ in sign and magnitude of the components, indicating a change in both speed and direction.

Key concepts

Magnitude and Direction

When discussing motion, especially in physics, the concepts of magnitude and direction are vital. Magnitude refers to the size or level of a quantity, without any regard to its direction. In the context of acceleration, magnitude gives us a measure of how quickly an object is speeding up or slowing down.

For acceleration, we find the magnitude using the Pythagorean theorem. In our exercise, after calculating the components of the average acceleration in the x and y directions, we used these to find the magnitude of average acceleration as \( a_{\text{avg}} = \sqrt{(-8.67)^2 + (-2.33)^2} \), leading to approximately \( 8.98 \, \text{m/s}^2 \).

Direction, on the other hand, tells us the line or course along which something moves. It is described by the angle it forms with a reference line, often the x-axis. Using trigonometric functions, specifically the arctangent function, we determined the direction of the average acceleration as \( \theta = \arctan \left( \frac{-2.33}{-8.67} \right) \). This calculation equated to an angle of \( 15.1^\circ \) above the negative x-axis. This direction is crucial in understanding the true path of the accelerating motion.

Velocity Vectors

Velocity vectors are a graphical representation of how fast an object moves and in which direction.

They are expressed with both magnitude (speed) and direction and can be broken down into components along the x and y axes, making calculations easier. In our example, the jet plane's velocity at time \( t_1 \) is captured by the vector with components \( v_{x1} = 90 \, \text{m/s} \) and \( v_{y1} = 110 \, \text{m/s} \), which points in the positive direction on both axes.

Use arrows on a graph to depict these vectors, where the length of the arrow shows the speed and the direction represents the direction of movement. At time \( t_2 \), the plane's velocity is represented by \( v_{x2} = -170 \, \text{m/s} \) and \( v_{y2} = 40 \, \text{m/s} \), showing a significant change.

By comparing these vectors, we see the plane changes direction sharply, as indicated by a negative x-component and a reduced y-component. This visual method helps in assessing how the plane's motion has varied over time.

Change in Velocity

The change in velocity helps us understand how an object's speed and direction transform over time. It forms the basis for calculating acceleration.

Change in velocity is defined as the difference between the initial and final velocity vectors. For our scenario, we calculated the changes in velocity components as \( \Delta v_x = -260 \, \text{m/s} \) and \( \Delta v_y = -70 \, \text{m/s} \), considering the times \( t_1 \) and \( t_2 \).

This change signifies that the jet plane not only slows down considerably in the x-direction but also slightly in the y-direction. Understanding this shift is crucial because it provides insights into the forces acting on the jet over the time interval.

By knowing these changes, we can further compute the average acceleration, giving us deeper insights into the jet plane's maneuver. Such calculations are fundamental, especially in aerodynamics, where changes in speed and direction are continuous and vital for navigation and control.