A GPS tracking device is placed in a police dog to monitor its whereabouts relat

Question

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 degrees north of east. At time t2= 57 min, the dog's displacement from the station is 2.0 km, 75 degrees north of east. Find the magnitude and direction of the dog's average velocity between these two times.

Everything I have tried does not give me the same answer as the correct one which is 0.67 m/s, 21 degrees west of north. Please help with the steps!

Explanation / Answer

see if this helps: 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[(x2-x1)^2 + (y2-y1)^2]

The direction of the resultant, angle C is:

C = tan^-1[(y2-y1)/(x2-x1)]

(C is measured from the positive x axis)

***Other trigonometric methods are possible or a scale diagram. 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[(x2-x1)^2 + (y2-y1)^2]

The direction of the resultant, angle C is:

C = tan^-1[(y2-y1)/(x2-x1)]

(C is measured from the positive x axis)

***Other trigonometric methods are possible or a scale diagram.

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