A Siberian flying squirrel is observed wandering around near Irkutsk. First, the

Question

A Siberian flying squirrel is observed wandering around near Irkutsk. First, the squirrel travels through a displacement = 6.00 miles, directed 60° S of E. Then the squirrel undergoes a displacement D2= 10.0 miles, directed 700W of s. Thus, the squirrel undergoes a resultant displacement R-D+ D In the following, assume that the +y direction is defined as due North and that the +x direction is defined as due East, as shown in the figure 16. True or False: R D a. True b. False b. False and .jy 17. True or False: yare both negative? a. True 18. Displacement D has an y component-vector equal to a. -9.40 miles b. -9.40 miles. c. 3.42 miles.d.3.42 miles e. 3.42 miles. 19. The squirrel 's resultant displacement R has a magnitude of a. 16.0 miles. b. 15.01 miles. c. 10.7 miles. d. 21.0 miles. e. 12.0 miles. The squirrel's resultant displacement is directed a. due South-West. b. 53.4° Sof W. c. 36.6° W of N. d. 53.40 N of E. e. 36.6 S of w 20.

Explanation / Answer

D1 = 6 cos60 x^ - 6sin60y^

D2 = -10sin70x^ - 10cos70y^

R = D1 +D2 = 6 cos60 x^ - 6sin60y^ -10sin70x^ - 10cos70y^ = (6cos60 - 10sin70)x^ - (6sin60 + 10cos70)y^

|D1| = 6

|D2| = 10

|R| = sqrt((6cos60 - 10sin70)2 + (6sin60 + 10cos70)2) < 16 = |D1| + |D2|

16. False

17. D1y = - 6sin60

D2y = - 10cos70

True

18. D2y = - 10cos70 = -3.42 miles j^

19 |R| = sqrt((6cos60 - 10sin70)2 + (6sin60 + 10cos70)2) = 10.73 miles

20. as net displacement in x and y is negative (3rd quadrant)

angle = tan-1((6sin60 + 10cos70)/(10sin70 - 6cos60)) S of W = 53.4 S of W

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