physics
posted by Joe .
An electronic tracking device is placed on a police dog to monitor its whereabouts relative to the police station. At time t1 = 23 minutes, the dogs placement from the station is 1.2km, 33 degrees north of east. At time t2=57 minutes, the dog's displacement from the station is 2.0km, 75 degrees north of east. Find the magnitude and direction of the dog's average velocity between these two times.

To get the coordinates of each point you can use:
(x1, y1) > x1=r1*cosA, y1=r1*sinA
(r1 = 1.2km, A = 33deg)
(x2, y2) > x2=r2*cosB, y2=r2*sinB
(r2=2.0km, B = 75deg)
The displacement is:
d = sqrt[(x2x1)^2 + (y2y1)^2]
The direction of the resultant, angle C is:
C = tan^1[(y2y1)/(x2x1)]
(C is measured from the positive x axis)
***Other trigonometric methods are possible or a scale diagram.
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