The position of Bob\'s apartment is 5.1 km and 50 degrees east of north relative

Question

The position of Bob's apartment is 5.1 km and 50 degrees east of north relative to the center of the OSU campus. His first class of the day starts at 9: 00 AM and is held in a room located 0.85 km and 30 degrees east of south relative to the campus center. Take the unit vector i^as being in a easterly direction and the unit vector j^as being an northerly direction. Calculate the following: a) The position of both his apartment and the classroom in terms of the unit vectors. b) The straight-line distance between his apartment and the classroom. c) He can ride his bicycle at a speed of 20 km/hr and takes a straight-line (direct) route to class. At what time will he need to leave home to get to class on time? Supply the answer below and attach additional pages if you require them (which must be stapled to the cover sheet). To receive credit you must show all your working.

Explanation / Answer

a)

position of his apartment in verrctor notation:

r1 = 5.1*( sin(50 deg) i + cos(50 deg) j )

= ( 3.91 i + 3.3 j ) km

r2 = 0.85*( sin(30 deg) i - cos(30 deg) j)

= ( 0.43 i - 0.74 j ) km

b)

distance between his apartment and classroom:

D = sqrt ( (3.91 - 0.43)^2 + (3.3 + 0.74)^2 )

= 5.33 km

c)

Time taken to reach classroom:

t = distance/speed = 5.33/20

= 0.2665 h = 0.2665*60 minutes

= 16 minutes <---------- answer

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