Alarge box is being pushed by three separate people. the first person is pushing the box at a rate of 6 m/s directly north. the second person is pushing the box at a rate of 6 m/s directly to the northeast. the third person is pushing the box at a rate of 6 m/s directly east. determine the magnitude and direct of the resultant vector of he box.

We can find the east component of the velocity. v_east = 6 m/s + (6 m/s) (cos (45)) v_east = 6 m/s + 4.24 m/s v_east = 10.24 m/s We can find the north component of the velocity. v_north = 6 m/s + (6 m/s) (sin (45)) v_north = 6 m/s + 4.24 m/s v_north = 10.24 m/s We can find the magnitude of the velocity. v = sqrt { (v_east)^2 + (v_north)^2} v = sqrt { (10.24 m/s)^2 + (10.24 m/s)^2} v = 14.5 m/s We can find the angle theta north of east. tan(theta) = 10.24 / 10.24 theta = arctan (1) theta = 45 Note that this angle is directly northeast. The magnitude of the velocity is 14.5 m/s and the box is moving directly to the northeast.

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