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Chapter 3: Problem 5

A jet plane is flying at a constant altitude. At time \(t_{1}=0\) it has components of velocity \(v_{x}=90 \mathrm{m} / \mathrm{s}, v_{y}=110 \mathrm{m} / \mathrm{s}\) . At time \(t_{2}=30.0 \mathrm{s}\) the components are \(v_{x}=-170 \mathrm{m} / \mathrm{s}, v_{y}=40 \mathrm{m} / \mathrm{s}\) (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

Expert verified
The jets' velocity components change direction and magnitude. The average acceleration is \(8.98 \, \mathrm{m/s^2}\) at \(15.0^\circ\) from the negative x-axis.

Step by step solution

01

Understanding Initial and Final Velocities

At time \(t_1 = 0\), the velocity components of the jet are \(v_{x1} = 90\, \mathrm{m/s}\) and \(v_{y1} = 110\, \mathrm{m/s}\). At time \(t_2 = 30.0\, \mathrm{s}\), the velocity components change to \(v_{x2} = -170\, \mathrm{m/s}\) and \(v_{y2} = 40\, \mathrm{m/s}\). We need to find how these vectors differ by sketching them and analyzing their change in direction and magnitude.
02

Sketching Velocity Vectors

For vector at \(t_1\), plot the point using the components \(v_{x1} = 90\) along the x-axis and \(v_{y1} = 110\) along the y-axis. This creates a vector from the origin to (90, 110). For \(t_2\), the vector starts from the origin to (-170, 40). Note the changes: \(v_{x}\) went from positive to negative, and \(v_{y}\) decreased.
03

Calculating Average Acceleration Components

The average acceleration components \(a_x\) and \(a_y\) can be calculated using the formula: \[a_{x} = \frac{v_{x2} - v_{x1}}{t_2 - t_1}\quad \text{and} \quad a_{y} = \frac{v_{y2} - v_{y1}}{t_2 - t_1}\]Substituting the values, we get:\[a_{x} = \frac{-170 - 90}{30} = \frac{-260}{30} = -8.67\, \mathrm{m/s^2}\]\[a_{y} = \frac{40 - 110}{30} = \frac{-70}{30} = -2.33\, \mathrm{m/s^2}\]
04

Calculating Magnitude of Average Acceleration

The magnitude of the average acceleration \(a\) can be found using the Pythagorean theorem:\[a = \sqrt{a_x^2 + a_y^2}\]Substitute the calculated acceleration components:\[a = \sqrt{(-8.67)^2 + (-2.33)^2} = \sqrt{75.23 + 5.43} = \sqrt{80.66} \approx 8.98\, \mathrm{m/s^2}\]
05

Calculating Direction of Average Acceleration

The direction \(\theta\) of the average acceleration vector (counterclockwise from the positive x-axis) is found using the tangent ratio:\[\theta = \tan^{-1}\left(\frac{a_y}{a_x}\right) = \tan^{-1}\left(\frac{-2.33}{-8.67}\right) = \tan^{-1}(0.268)\]Calculating this gives \(\theta \approx 15.0^\circ\). Since both \(a_x\) and \(a_y\) are negative, this angle is actually with respect to the negative x-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Acceleration
Average acceleration is the rate at which velocity changes over time. In practical terms, it describes how quickly something speeds up or slows down. To compute this, we need both initial and final velocities along with the time interval over which the change occurs.

For our jet plane scenario, the initial velocity components are given at time \( t_1 = 0 \, \text{s} \), and the final components are known at \( t_2 = 30.0 \, \text{s} \). Average acceleration components are calculated using:
  • \( a_x = \frac{v_{x2} - v_{x1}}{t_2 - t_1} \)
  • \( a_y = \frac{v_{y2} - v_{y1}}{t_2 - t_1} \)
Substituting the given numbers, we find:
  • \( a_x = -8.67 \, \mathrm{m/s^2} \)
  • \( a_y = -2.33 \, \mathrm{m/s^2} \)
These negative values indicate a deceleration in both directions.
Velocity Components
Velocity is a vector quantity, which means it has both magnitude and direction. It can be broken down into components along the x and y axes. This helps in analyzing the movement in a two-dimensional plane.

In the given problem, at \( t_1 \), the velocity components are \( v_{x1} = 90 \, \mathrm{m/s} \) and \( v_{y1} = 110 \, \mathrm{m/s} \). At \( t_2 \), they change to \( v_{x2} = -170 \, \mathrm{m/s} \) and \( v_{y2} = 40 \, \mathrm{m/s} \).

This shows:
  • The x-component changes from positive to negative, indicating a reversal in the direction along the x-axis.
  • The y-component decreases, suggesting a slower motion in the y-direction.
These changes are critical in determining the jet's overall path and speed changes.
Vector Analysis
To fully understand the movement of the jet, it is essential to analyze the vectors involved. Vector analysis allows us to visualize and calculate the resultant effects of different forces and movements.

Vector components \((v_x, v_y)\) can be plotted and combined to visualize the overall direction and speed of a moving object. For the jet, plotting these vectors at \( t_1 \) and \( t_2 \) shows:
  • The initial vector points to the first quadrant, indicating forward and upward movement.
  • The final vector moves to the second quadrant, suggesting a leftward and still upward movement, but slower.
Using the Pythagorean theorem, we calculate the magnitude of velocity changes over time, aiding in understanding the dynamics of motion.
Jet Plane Motion
Jet plane motion involves understanding how high-speed objects navigate through airspace. In the context of physics, it's crucial to consider how rapidly changing speeds and directions affect overall motion.

In this exercise, a jet plane's changing velocity components provide insights into its movement patterns. At different times, various forces, such as thrust and drag, influence these components. This results in:
  • Changes in speed and direction as described by different velocity components at \( t_1 \) and \( t_2 \).
  • Causing the jet to decelerate in one direction while altering its path.
Overall, analyzing jet plane motion involves considering these complex interactions to predict future positions and paths.

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Most popular questions from this chapter

A particle moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$x(t)=R(\omega t-\sin \omega t) \quad y(t)=R(1-\cos \omega t)$$ where \(R\) and \(\omega\) are constants. (a) Sketch the trajectory of the particle. This is the trajectory of a point on the rim of a wheel that is rolling at a constant speed on a horizontal surface. The curve traced out by such a point as it moves through space. is called a cvcloid.) (b) Determine the velocity components and the acceleration components of the particle at any time \(t\) (c) At which times is the particle momentarily at rest? What are the coordinates of the particle at these times? What are the magnitude and direction of the acceleration at these times? (d) Does the magnitude of the acceleration depend on time? Compare to uniform circular motion.

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A man stands on the roof of a 15.0 -m-tall building and throws a rock with a velocity of magnitude 30.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. You can ignore air resistance. Calculate (a) the maximum height above the roof reached by the rock; (b) the magnitude of the velocity of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t\) , \(v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

An elevator is moving upward at a constant speed of2.50 \(\mathrm{m} / \mathrm{s} .\) A bolt in the elevator ceiling 3.00 \(\mathrm{m}\) above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings of the building? (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?

A rock is thrown from the roof of a building with a velocity \(v_{0}\) at an angle of \(\alpha_{0}\) from the horizontal. The building has height \(h .\) You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground, and show that this speed is independent of \(\alpha_{0}\) .
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