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.

Question

Mareli Vargas

Physics

16 October, 10:22

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.

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Pugsley · Answer 1

Let east be towards X-axis ant north be Y-axis. Let initial position of Dog be at

A. O be the police station (centre). Vector OA can be written as follows

OA = 1.2 Cos 33 + 1.2 Sin 33

O B = 2 Cos 75 + 2 Sin 75.

Displacement A → B = OB - OA = 2Cos75i + 2 Sin 75j - 1.2 Cos 33i - 1.2 Sin 33j

= 2 x. 2588 i + 2x. 966j - 1.2 x. 8387i - 1.2 x. 5446j =.5176i + 1.932 j - 1.0064i -.65356j

= -.4888 i + 1.27844j

Magnitude of displacement = √ (.4888) ² + (1.27844) ²

= 1.405 km

Average velocity = 1.405 / 57-23 km / min = 1.405 / 34 x 60 = 2.48 km/h

angle with x-axis (east towards north) ∅

Tan∅ = - 1.27844/.4888 = - 2.615

∅ = - 69° or 111° towards north from east or 21° towards west from north.