It moves towards the East a distance d1=6km and then heads towards the Northwest a distance of d2=4km Determine the resulting displacement D in magnitude and direction

D = sqrt((6 km)^2 + (4 km)^2) = 7.21 km To determine the direction of the displacement vector, we can use trigonometry. The angle between the horizontal component and the displacement vector is given by: theta = arctan(4 km / 6 km) = 33.69 degrees Therefore, the resulting displacement D has a magnitude of 7.21 km and a direction of 33.69 degrees North of East.