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Chapter 4: Problem 8

A plane flies \(483 \mathrm{~km}\) east from city \(A\) to city \(B\) in \(45.0 \mathrm{~min}\) and then \(966 \mathrm{~km}\) south from city \(B\) to city \(C\) in \(1.50 \mathrm{~h}\). For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed?

Short Answer

Expert verified
Displacement is 1084 km at 63.4掳 S of E; average velocity is 481 km/h at the same direction; average speed is 644 km/h.

Step by step solution

01

Understand the Displacement

We have two legs of the journey: an eastward displacement from city A to city B of \(483\;km\) and a southward displacement to city C of \(966\;km\). To find the total displacement, we consider these as vectors in a coordinate system, with east as the positive x-axis and south as the negative y-axis.
02

Calculate Total Displacement Magnitude

The total displacement is the vector sum of the two legs. Using the Pythagorean theorem, the magnitude of the displacement \(d\) is given by: \[ d = \sqrt{(483)^2 + (966)^2}\] Calculate \(d\).
03

Determine Direction of Displacement

The direction \(\theta\) of the displacement vector from city A to city C is given by an angle south of east. Using the tangent function: \[ \theta = \tan^{-1}\left(\frac{966}{483}\right)\] Compute \(\theta\).
04

Find Total Time of the Journey

The total time taken is the sum of the time for each leg: Eastward: \(45.0\;min = 0.75\;hours\) Southward: \(1.50\;hours\). So the total time \(t\) is \(0.75 + 1.50 = 2.25\;hours\).
05

Calculate Average Velocity's Magnitude

The average velocity \(v_{avg}\) is the total displacement divided by the total time. Using the magnitude from Step 2: \[ v_{avg} = \frac{d}{2.25}\] Substitute \(d\) and calculate \(v_{avg}\).
06

Determine Direction of Average Velocity

The direction of the average velocity is the same as the displacement direction \(\theta\) calculated in Step 3.
07

Compute Average Speed

Average speed is the total distance traveled divided by the total time: Total distance = \(483\;km + 966\;km = 1449\;km\). \[ \text{Average speed} = \frac{1449}{2.25}\] Calculate the average speed.

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Key Concepts

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

Average Velocity
Average velocity is an important concept in physics that helps to understand how fast an object is moving in a particular direction. It is defined as the total displacement divided by the total time taken for the journey.
  • Total Displacement: This is calculated using the Pythagorean theorem, where the displacement is the straight-line distance between the starting and the ending points of the journey.

  • Total Time: In this context, it is the sum of the time taken to travel each segment of the plane's route.

Thus, to find the average velocity, we take the magnitude of the displacement and divide it by the total time. Since displacement is a vector, average velocity also has both magnitude and direction. In this particular exercise, the average velocity direction is the same as the direction of the overall displacement.
Pythagorean Theorem
The Pythagorean theorem is a mathematical principle used to find the length of the sides in a right-angled triangle. It is a foundational concept in trigonometry and is essential for calculating displacement in problems involving perpendicular directions.
  • The formula is given by: \[c = \sqrt{a^2 + b^2}\]where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides.

  • In this exercise, we use the Pythagorean theorem to calculate the magnitude of the total displacement from city A to city C, treating the east and south travels as perpendicular components.

By employing this theorem, students can easily determine how the plane's path transforms into a direct line of travel across two dimensions. This helps to visualize how seemingly complex motions can be distilled into simpler calculations.
Direction of Motion
Direction of motion is essential in understanding the path an object follows. It is described using angles and often involves trigonometric functions like sine, cosine, and tangent. In this exercise, direction is crucial in evaluating the plane's trajectory from city A to city C.
  • The direction's angle helps denote the orientation of motion, here calculated as south of east using:\[\theta = \tan^{-1}\left(\frac{\text{distance south}}{\text{distance east}}\right)\]

  • This calculation gives the angle \(\theta\), providing understanding of how the plane moves relative to a horizontal reference line.

Understanding direction not only assists in solving physics problems but also in navigation and coordinating movements in everyday life or advanced fields like aeronautics.
Average Speed
Average speed is a fundamental concept that describes how fast an object is traveling regardless of direction. Unlike average velocity, it only considers the total path distance and is always scalar.
  • The average speed is determined by dividing the total distance traveled by the total time taken:
  • Total distance in the exercise is calculated as the sum of both legs of the journey (east and south travels).

  • Total time is the entire duration spent on the journey.

The average speed gives insight into the plane's overall performance and efficiency during its trip, illustrating how it covers space over time without considering directional changes. Understanding the distinction between speed and velocity is key to mastering many motion-related challenges in physics.

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Most popular questions from this chapter

A sprinter running on a circular track has a velocity of constant magnitude \(9.20 \mathrm{~m} / \mathrm{s}\) and a centripetal acceleration of magnitude \(3.80 \mathrm{~m} / \mathrm{s}^{2}\). What are (a) the track radius and (b) the period of the circular motion?

A woman who can row a boat at \(6.4 \mathrm{~km} / \mathrm{h}\) in still water faces a long, straight river with a width of \(6.4 \mathrm{~km}\) and a current of \(3.2 \mathrm{~km} / \mathrm{h}\). Let \(\hat{\text { i }}\) point directly across the river and \(\hat{\mathrm{j}}\) point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to 卯 must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rows \(3.2 \mathrm{~km}\) down the river and then back to her starting point? (d) How long if she rows \(3.2 \mathrm{~km} u p\) the river and then back to her starting point? (e) At what angle to \(\hat{\mathbf{i}}\) should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

A cart is propelled over an \(x y\) plane with acceleration components \(a_{x}=4.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(a_{y}=-2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Its initial velocity has components \(v_{0 x}=8.0 \mathrm{~m} / \mathrm{s}\) and \(v_{0 y}=12 \mathrm{~m} / \mathrm{s}\). In unit-vector notation, what is the velocity of the cart when it reaches its greatest \(y\) coordinate? ..18 A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\).

An electron having an initial horizontal velocity of magnitude \(1.00 \times 10^{9} \mathrm{~cm} / \mathrm{s}\) travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of \(2.00 \mathrm{~cm}\) and has a constant downward acceleration of magnitude \(1.00 \times 10^{17} \mathrm{~cm} / \mathrm{s}^{2}\) due to the charged plates. Find (a) the time the electron takes to travel the \(2.00 \mathrm{~cm},(\mathrm{~b})\) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

A particle leaves the origin with an initial velocity \(\vec{v}=(3.00 \hat{\mathrm{i}}) \mathrm{m} / \mathrm{s}\) and a constant acceleration \(\vec{a}=(-1.00 \hat{\mathrm{i}}-\) \(0.500 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}\). When it reaches its maximum \(x\) coordinate, what are its (a) velocity and (b) position vector?
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