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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.

  • physics -

    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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