Chapter 3: Problem 5
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
Step by step solution
Identify the Components of Velocity
Calculate the Change in Velocity Components
Calculate the Components of Average Acceleration
Calculate the Magnitude of the Average Acceleration
Determine the Direction of the Average Acceleration
Sketch and Compare the Velocity Vectors
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude and Direction
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
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
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.