We define our coordinate system such that the positive X-axis points East and the positive Y-axis points North.

Here's a step-by-step solution to the problem: We define our coordinate system such that the positive X-axis points East and the positive Y-axis points North. Part (i): The components of each displacement Displacement 1 (d_1): 4.0 m South West South West means the angle is 225^ from the positive X-axis, or 45^ South of West. d_1x = 4.0 (225^) = 4.0 (-sqrt(2)2) = -2sqrt(2) m ≈ -2.83 m d_1y = 4.0 (225^) = 4.0 (-sqrt(2)2) = -2sqrt(2) m ≈ -2.83 m So, d_1 = (-2sqrt(2) m, -2sqrt(2) m) Displacement 2 (d_2): 5.0 m East This displacement is entirely along the positive X-axis. d_2x = 5.0 m d_2y = 0 m So, d_2 = (5.0 m, 0 m) Displacement 3 (d_3): 6.0 m in a direction 30° E of N "30° E of N" means 30^ from the North axis towards the East. This corresponds to an angle of 90^ - 30^ = 60^ from the positive X-axis. d_3x = 6.0 (60^) = 6.0 ((1)/(2)) = 3.0 m d_3y = 6.0 (60^) = 6.0 (sqrt(3)2) = 3sqrt(3) m ≈ 5.20 m So, d_3 = (3.0 m, 3sqrt(3) m) Part (ii): The components of the resultant displacement Let the resultant displacement be R. We sum the respective components: R_x = d_1x + d_2x + d_3x R_x = -2sqrt(2) + 5.0 + 3.0 = (8.0 - 2sqrt(2)) m R_y = d_1y + d_2y + d_3y R_y = -2sqrt(2) + 0 + 3sqrt(3) = (3sqrt(3) - 2sqrt(2)) m Numerically: R_x ≈ 8.0 - 2(1.414) = 8.0 - 2.828 = 5.172 m R_y ≈ 3(1.732) - 2(1.414) = 5.196 - 2.828 = 2.368 m The components of the resultant displacement are (5.17 m, 2.37 m). Part (iii): The magnitude and direction of the resultant displacement The magnitude of the resultant displacement |R| is: |R| = sqrt(R_x^2 + R_y^2) |R| = sqrt((5.172)^2 + (2.368)^2) |R| = sqrt(26.75 + 5.61) = sqrt(32.36) ≈ 5.69 m The direction is given by: = ((R_y)/(R_x)) = ((2.368)/(5.172)) = (0.4578) ≈ 24.6^ Since both R_x and R_y are positive, the resultant displacement is in the first quadrant. The magnitude of the resultant displacement is 5.69 m and its direction is 24.6^ North of East. Part (iv): The displacement that would be required to bring the rat back to the starting point To bring the rat back to the starting point, the required displacement must be equal in magnitude and opposite in direction to the resultant displacement. Let this be D_return. D_return = -R D_return, x = -R_x = -(8.0 - 2sqrt(2)) m ≈ -5.17 m D_return, y = -R_y = -(3sqrt(3) - 2sqrt(2)) m ≈ -2.37 m The magnitude of this displacement is the same as |R|, which is approximately 5.69 m. To find the direction, we note that both components are negative, placing the displacement in the third quadrant (South-West). The angle relative to the negative X-axis (West) is: = (|D_return, y||D_return, x|) = ((2.368)/(5.172)) ≈ 24.6^ The displacement required to bring the rat back to the starting point has a magnitude of $5.69