Sample Problem 3.03 Searching through a hedge maze A hedge maze is a maze formed

Question

Sample Problem 3.03 Searching through a hedge maze A hedge maze is a maze formed by tall rows of hedge. Calculations: To evaluate Eqs. 3-16 and 3-17, we find the x and After entering, you search for the center point and then y components of each displacement. As an example, the com- for the exit. Figure 3-16a shows the entrance to such a ponents for the first displacement are shown in Fig. 3-16c. We maze and the first two choices we make at the junctions draw similar diagrams for the other two displacements and we encounter in moving from point i to point c. We un- then we apply the x part of Eq. 3-5 to each displacement, using dergo three displacements as indicated in the overhead angles relative to the positive direction of the x axis: view of Fig. 3-16b: d,-(6.00 m) cos 40°-460 m di = 6.00 m d2 = 8.00 m d3 = 5.00 m | = 400 e=300 3 = 0°, = (8.00 m) cos (-60°) = 4.00 m d3,-(5.00 m) cos 0°-5.00 m. Equation 3-16 then gives us dnet where the last segment is parallel to the superimposed x axis. When we reach point c, what are the magnitude and angle of our net displacement d net from point i? +4.60 m 4.00 m 5.00 m = 13.60 m. Similarly, to evaluate Eq.3-17, we apply the y part of Eq. 3-5 to each displacement KEY IDEAS (1) To find the net displacement dact, we need to sum the three individual displacement vectors: diy- (6.00 m) sin 40° 3.86 m dy = (8.00 m) sin (-60°)--6.93 m dsy- (5.00 m) sin 0°0m. dnet= d, + d2 + d3 (2) To do this, we first evaluate this sum for the x compo- nents alone, Equation 3-17 then gives us (3-16) dnet.v = +3.86 m-6.93 m 0 m =-3.07 m. and then the y components alone, Next we use these components of det to construct the vec tor as shown in Fig. 3-16d: the components are in a head-to- (3) Finally, we construct dnet from its x and y components tail arrangement and form the legs of a right triangle, and Three vectors First vector Net ector Figure 3-16 (a) Three displacements through a hedge maze. (b) The displacement vectors. (c) The first displacement vector and its components. (d) The net displacement vector and its components

Explanation / Answer

As for the vertical component of d2 the cos(theta 2) must be multiplied but you are multiplying the sine function so the (90-theta 2) will be used.

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