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
Here's how to determine the distance between the two insects after 5 seconds: Step 1: Determine the position vector of insect A after 5 s. Assuming both insects start at the origin (0,0,0) at t=0, the position vector R_A at time t is given by R_A = V_A t. Given V_A = i + 4j + 3k m/s and t = 5 s. R_A = (i + 4j + 3k) × 5 R_A = 5i + 20j + 15k \, m Step 2: Determine the position vector of insect B after 5 s. Similarly, the position vector R_B at time t is given by R_B = V_B t. Given V_B = 4i + 2j - 4k m/s and t = 5 s. R_B = (4i + 2j - 4k) × 5 R_B = 20i + 10j - 20k \, m Step 3: Determine the relative position vector between A and B after 5 s. The relative position vector R_AB (position of B relative to A) is R_AB = R_B - R_A. R_AB = (20i + 10j - 20k) - (5i + 20j + 15k) R_AB = (20 - 5)i + (10 - 20)j + (-20 - 15)k R_AB = 15i - 10j - 35k \, m Step 4: Calculate the distance apart. The distance apart is the magnitude of the relative position vector R_AB. |R_AB| = sqrt((15)^2 + (-10)^2 + (-35)^2) |R_AB| = sqrt(225 + 100 + 1225) |R_AB| = sqrt(1550) |R_AB| ≈ 39.3700 \, m Comparing this value with the given options, the closest match is 39.37m. The final answer is 39.37m. That's 2 down. 3 left today — send the next one.