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.
Find the following for path C 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 C, it moves 1 block upward, 5 blocks to the right, 2 blocks downward, 1 block to the left, 1 block upward, and 3 blocks to the left. So, the total distance is
It can be seen from the figure that the end of path C is just one block to the right from the starting point. Therefore, the magnitude of the displacement is
(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R’=A−B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R”=B−A=−R’ ). Show that this is the case.
Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)
Solution:
Consider Figure 3-6A below with B drawn first before A.
Compute for the value of angle β by adding 20° and the complement of 40°. This is by simple geometry.
Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. 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 3.54, then this problem finds their sum R=A+B.)
Solution:
Consider Figure 3.5A shown below.
Before we can use cosine law to solve for the magnitude of R, we need to solve for the interior angle 𝛽 first. The value of 𝛽 can be calculated by inspecting the figure and use simple knowledge on geometry. It is equal to the sum of 20° and the complement of 40°. That is
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 \vec{A} and \vec{B} , as in Figure 3.53, then this problem asks you to find their sum \vec{R}=\vec{A}+\vec{B} .)
Solution:
Consider Figure 3.54A.
The resultant of the two vectors \vec{A} and \vec{B} is labeled \vec{R}. This \vec{R} is directed \theta ^{\circ} from the x-axis.
We shall use the right triangle formed to solve for the unknowns.
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