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Chapter 3: Problem 18

A small map shows Atlanta to be 730 miles in a direction \(5^{\circ}\) north of east from Dallas. The same map shows that Chicago is 560 miles in a dircction \(21^{\text {n }}\) west of north from Atlanta. Assume a flat Earth and use the given information to find the displacement from Dallas to Chicago.

Short Answer

Expert verified
The displacement from Dallas to Chicago is 595 miles.

Step by step solution

01

Diagram and Vector Decomposition for Atlanta from Dallas

First, draw a rough sketch of the situation. Then find the coordinates for Atlanta in reference to Dallas by resolving the vector. Convert the polar coordinates (730 miles, \(5o\)), into Cartesian coordinates to get the east and north components of the Dallas-Atlanta vector: \(\Delta x = R \cos \theta = 730 \cos 5 = 727.66\) miles and \(\Delta y = R \sin \theta = 730 \sin 5 = 63.56\) miles. So, the position of Atlanta in reference to Dallas is (727.66, 63.56).
02

Diagram and Vector Decomposition for Chicago from Atlanta

Carry out a similar process for Chicago in reference to Atlanta, noting the direction change. Chicago鈥檚 x (westward) and y (northward) displacements are \(\Delta x = R \cos \theta = 560 \cos 21 = -192.99\) miles and \(\Delta y = R \sin \theta = 560 \sin 21 = 201.54\) miles. So, the position of Chicago in reference to Atlanta is (-192.99, 201.54).
03

Summing the Vectors

To find Dallas-Chicago distance, find the vector sum of the Dallas-Atlanta vector and the Atlanta-Chicago vector. This means adding the x and y coordinates of each vector together: \(\Delta x = 727.66 - 192.99 = 534.67\) miles and \(\Delta y = 63.56 + 201.54 = 265.10\) miles. This will provide the Cartesian coordinates (534.67, 265.10) for the Dallas-Chicago vector.
04

Calculating the Displacement

Finally, calculate the magnitude of the Dallas-Chicago vector using the Pythagorean Theorem. This will give the direct distance (displacement) from Dallas to Chicago: \( R = \sqrt{\Delta x^2 + \Delta y^2} = \sqrt{(534.67^2 + 265.10^2)} = 595\) miles.

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Key Concepts

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

Displacement Vector
Understanding displacement vectors is crucial in finding the shortest path between two points, such as Dallas and Chicago. A displacement vector is a quantity that reflects both distance and direction between two locations. In our exercise, the displacement vector from Dallas to Chicago takes into account the individual journeys from Dallas to Atlanta, and then from Atlanta to Chicago.
  • It starts with decomposing each leg of the journey into vector components using specific angles and distances.
  • The vector sum of these components gives the overall displacement vector, representing the direct route one would take when flying or driving directly.
By using vector decomposition, we simplify complex navigation problems into manageable calculations. Each segment's direction and distance are crucial to accurately determine the end displacement.
Cartesian Coordinates
Cartesian coordinates provide a way to map points in a two-dimensional plane, allowing us to analyze displacement vectors more comprehensively. By translating distances and angles into Cartesian coordinates, we can work with straightforward x (horizontal) and y (vertical) components. In the problem, the positions of Atlanta and Chicago relative to Dallas are expressed using these coordinates.
  • The conversion from polar coordinates, which use radial distance and angle, to Cartesian coordinates involves trigonometric functions: cosine for the x-component and sine for the y-component.
  • This conversion simplifies the addition or subtraction of vectors, as it allows for direct algebraic operations on the x and y components.
Thus, Cartesian coordinates serve as an efficient framework for dealing with displacement vectors, making complex navigation questions more solvable.
Pythagorean Theorem
The Pythagorean theorem is essential for determining the magnitude of resultant vectors after decomposition and summation. It applies perfectly in calculating the direct distance between two points, such as between Dallas and Chicago.
  • The theorem states that the square of the hypotenuse (longest side of a right triangle) is equal to the sum of the squares of the other two sides.
  • For vectors, this means using the formula \( R = \sqrt{\Delta x^2 + \Delta y^2} \) to find the displacement magnitude, where \( \Delta x \) and \( \Delta y \) are the respective calculated x and y components of the vector.
This theorem allows for the transformation of the Cartesian coordinate components back into a meaningful measurement of distance鈥攁 crucial step to completing vector analysis.

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Most popular questions from this chapter

A hunter wishes to cross a river that is \(1.5 \mathrm{~km}\) wide and flows with a speed of \(5.0 \mathrm{~km} / \mathrm{h}\) parallel to its banks. The hunter uses a small powerboat that moves at a maximum speed of \(12 \mathrm{~km} / \mathrm{h}\) with respect to the water. What is the minimum time necessary for crossing?

A rocket is launched at an angle of \(53.0^{\circ}\) above the horizontal with an initial speed of \(100 \mathrm{~m} / \mathrm{s}\). The rocket moves for \(3.00 \mathrm{~s}\) along its initial line of motion with an acceleration of \(30.0 \mathrm{~m} / \mathrm{s}^{2}\). At this time, its engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and \((c)\) its horizontal range.

A home run is hit in such a way that the baseball just clears a wall \(21 \mathrm{~m}\) high, located \(130 \mathrm{~m}\) from home plate. The ball is hit at an angle of \(35^{\circ}\) to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall. and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height of \(1.0 \mathrm{~m}\) above the ground.)

A jogger runs \(100 \mathrm{~m}\) due west, then changes direction for the second leg of the run. At the end of the run, she is 175 \(\mathrm{m}\) away from the starting point at an angle of \(15.0^{\circ}\) north of west. What were the direction and length of her second displacement? Use graphical techniques.

The record distance in the sport of throwing cowpats is \(81.1 \mathrm{~m}\). This record toss was set by Steve Urner of the United States in 1981 . Assuming the initial launch angle was \(45^{\circ}\) and neglecting air resistance, determine (a) the initial speed of the projectile and (b) the total time the projectile was in flight. (c) Qualitatively, how would the answers change if the launch angle were greater than \(45^{\circ}\) ? Explain.
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