how do you calculate this? A GPS tracking device is placed in a police dog to mo
ID: 1597712 • Letter: H
Question
how do you calculate this?
A GPS tracking device is placed in a police dog to monitor its whereabouts relative to the police station. At time t1 = 23 min, the dog's displacement from the station is 1.2 km, 33° north of east. At time t2 = 57 min, the dog's displacement from the station is 2.0 km, 75° north of east. Find the magnitude and direction of the dog's average velocity between these two times. supposedly ppl say the answer is D
A) 0.52 m/s, 88° north of east B) 1.4 m/s, 31° west of north C) 1.6 m/s, 42° north of east D) 0.67 m/s, 21° west of north E) 0.80 m/s, 42° west of north
Explanation / Answer
(D) 0.67 m/s, 21° west of north
Distance moved till t1 = 23 min:
horizontal distance = 1.2 cos (33)
vertical distance = 1.2 sin (33)
Distance moved till t2 = 57 min:
horizontal distance = 2 cos (75)
vertical distance = 2 sin (75)
Distance moved between t1 and t2:
horizontal distance = 1.2 cos (33) - 2 cos (75) = 0.49Km
vertical distance = 1.2 sin (33) - 2 sin (75) = 1.28Km
total displacement = sqrt (0.49^2 + 1.28^2)
velocity = displacement/(57-23) = 0.67m/s
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