This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Define the velocities in component form. Let the positive x-axis be East and the positive y-axis be North. The velocity of the torpedo, v_T, is 32 m/s NW. NW means 45^ North of West. This corresponds to an angle of 135^ from the positive x-axis. v_T = (32 135^) i + (32 135^) j v_T = (32 × -sqrt(2)2) i + (32 × sqrt(2)2) j v_T = -16sqrt(2) i + 16sqrt(2) j m/s Numerically: v_Tx ≈ -16 × 1.4142 = -22.627 m/s v_Ty ≈ 16 × 1.4142 = 22.627 m/s The velocity of the freighter, v_F, is 10 m/s 60^ E of S. This means 60^ East from the South direction. South is -90^ or 270^. So the angle from the positive x-axis is 270^ + 60^ = 330^, or equivalently -30^. v_F = (10 (-30^)) i + (10 (-30^)) j v_F = (10 × sqrt(3)2) i + (10 × -(1)/(2)) j v_F = 5sqrt(3) i - 5 j m/s Numerically: v_Fx ≈ 5 × 1.7321 = 8.660 m/s v_Fy = -5 m/s Step 2: Calculate the relative velocity of the torpedo with respect to the freighter, v_T/F. The relative velocity is given by v_T/F = v_T - v_F. v_T/F = (-16sqrt(2) i + 16sqrt(2) j) - (5sqrt(3) i - 5 j) v_T/F = (-16sqrt(2) - 5sqrt(3)) i + (16\