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: Write down the given velocity vectors. The velocity of insect A is V_A = i + 4j + 3k m/s. The velocity of insect B is V_B = 4i + 2j - 4k m/s. Step 2: Calculate the dot product of the velocity vectors to show they fly at a right angle. If two vectors are perpendicular, their dot product is zero. V_A · V_B = (1)(4) + (4)(2) + (3)(-4) V_A · V_B = 4 + 8 - 12 V_A · V_B = 12 - 12 V_A · V_B = 0 Step 3: Conclude that the insects fly at a right angle. Since the dot product of V_A and V_B is zero, the velocity vectors are perpendicular. Therefore, the insects fly at a right angle to each other. Step 4: Determine the relative velocity vector. The relative velocity of insect B with respect to insect A is V_BA = V_B - V_A. V_BA = (4i + 2j - 4k) - (i + 4j + 3k) V_BA = (4-1)i + (2-4)j + (-4-3)k V_BA = 3i - 2j - 7k m/s Step 5: Calculate the relative displacement vector after 5 seconds. Assuming both insects start at the origin (0,0,0) at t=0, their relative displacement after time t is r_BA(t) = V_BA t. For t = 5 s: r_BA(5) = (3i - 2j - 7k) × 5 r_BA(5) = 15i - 10j - 35k m Step 6: Calculate the magnitude of the relative displacement vector to find the distance apart. The distance apart is the magnitude of r_BA(5). D = |r_BA(5)| = sqrt((15)^2 + (-10)^2 + (-35)^2) D = sqrt(225 + 100 + 1225) D = sqrt(1550) D ≈ 39.3700 m The distance apart after 5 seconds is: 39.37 m What's next?