You drive 7.50 km in a straight line in a direction 15º east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.
Solution:
Part A
Consider the illustration shown.
Let DE be the east component of the distance, and DNbe the north component of the distance.
Repeat Exercise 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.
Solution:
Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is
Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.58, then this problem asks you to find their sum R=A+B.)
Solution:
Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is
Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.57.
Solution:
Consider the following figure.
Using the right triangle formed, we can solve for the east component and the north component. Let SE be the east component and SN be the north component of S.
Find the following for path D in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.
Solution:
Part A
Looking at path D, we can see that it moves 2 blocks downward, 6 blocks to the right, 4 blocks upward, and 1 block to the left. Thus, the total distance of path D is
Looking at the initial and final position of path D, the final position is 5 blocks to the right or 600 meters to the right of the initial position, and 2 blocks or 240 meters upward from the initial position. Refer to the figure below.
Using the right triangle, we can solve for the displacement using the Pythagorean Theorem.
Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.61. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.
Solution:
The pilot’s displacement is characterized by 2 vectors, A and B, as depicted in Figure 3.61. To determine her total displacement R from the starting point, we need to add the two given vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:
Vector
x-component
y-component
A
40\:\cos 60^{\circ} =20\:\text{km}
40\:\sin 60^{\circ} =34.6410\:\text{km}
B
30\:\cos 15^{\circ} =28.9778\:\text{km}
30\:\sin 15^{\circ} =7.7646\:\text{km}
Sum
48.9778\: \text{km}
42.4056 \:\text{km}
The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. These are also the x and y components of the resultant vector.
To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:
Therefore, the pilot’s resultant displacement is about 64.8 km directed 40.9° North of East from the starting island.
Discussion:
If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will travel more to the east than without the wind. If the wind speed is greater than the speed of the plane, then it is no longer possible for the plane to travel to the northeast, it will end up traveling southeast.
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?
Solution:
Gilligan’s displacement is characterized by 7 vectors. To determine his final position relative to the starting point, we simply need to add the vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:
Vector
X-Component
Y-Component
(1)
-2.5\:\cos 45^{\circ} =-1.7678\:\text{km}
+2.5\:\sin 45^{\circ} =+1.7678\:\text{km}
(2)
+4.70\:\cos 60^{\circ} =+2.3500\:\text{km}
-4.70\:\sin 60^{\circ} =-4.0703\:\text{km}
(3)
-1.30\:\cos 25^{\circ} =-1.1782\:\text{km}
-1.30\:\sin 25^{\circ} =-0.5494\:\text{km}
(4)
+5.1000\:\text{km}
0
(5)
+1.70\:\sin 5^{\circ} =+0.1482\:\text{km}
+1.70\:\cos 5^{\circ} =+1.6935\:\text{km}
(6)
-7.20\:\cos 55^{\circ} =-4.1298\:\text{km}
-7.20\:\sin 55^{\circ} =-5.8979\:\text{km}
(7)
+2.80\:\cos 10^{\circ} =+2.7575\:\text{km}
+2.80\:\sin 10^{\circ} =+0.4862\:\text{km}
Sum
3.2799\:\text{km}
-6.5701\:\text{km}
The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.
To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:
Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A — that is, finding R’=A – B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A=B+C . Is that consistent with your result?)
Solution:
Part A
From the given statement, you first walk 18.0 m straight west and then 25.0 straight south. These vectors are represented by the graph shown below.
To solve for the resultant, we simply need to use the Pythagorean theorem to solve for the hypotenuse of the right triangle formed. That is,
So, we have a right triangle with legs 25.0 m and 18.0 m. We are tasked to solve for the value of R, and the angle θ for the compass direction. The value of R can be solved using the Pythagorean Theorem as in Part A.
To solve for the angle, θ, we shall use the tangent function.
Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.
Solution:
So, we are given the two vectors shown below.
If we use the graphical method of adding vectors, we can join the two vectors using head-tail addition and come up with the following:
The resultant is drawn from the tail of the first vectors (the origin) to the head of the last vector. The resultant is shown in red in the figure below.
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