Agps 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.

Explanation:

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.

c, 1a hope this

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