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
Welcome back grant — missed you this week. To solve this problem, we need to consider the velocities as vectors. The man wants his resultant velocity to be directly across the river, perpendicular to the river flow. Let v_m be the man's speed in still water, v_r be the river's speed, and v_res be the man's resultant speed across the river. Given: v_m = 6 km/hr v_r = 4 km/hr (eastwards) River width d = 5 km Step 1: Determine the direction the man must set off. To land directly opposite his starting point, the man must row upstream at an angle such that the component of his velocity against the river flow cancels out the river's velocity. This forms a right-angled triangle with v_m as the hypotenuse, v_r as the side opposite the angle of setting off upstream, and v_res as the side adjacent to this angle. Let be the angle upstream from the direction perpendicular to the river flow. Using trigonometry: = (v_r)/(v_m) = 4 km/hr6 km/hr = (2)/(3) To find the angle : = ((2)/(3)) ≈ 41.81^ The man must set off at an angle of 41.81^ upstream from the direction perpendicular to the river flow. If the river flows east, he must row 41.81^ west of north (assuming he wants to cross north). Step 2: Calculate the man's effective velocity across the river. The effective velocity across the river, v_res, is the component of the man's velocity perpendicular to the river flow. Using the Pythagorean theorem or the cosine of the angle: v_res = sqrt(v_m^2 - v_r^2) v_res = sqrt((6 km/hr))^2 - (4 km/hr)^2 v_res = sqrt(36 km)^2/hr^2 - 16 km^2/hr^2 v_res = sqrt(20 km)^2/hr^2 v_res = sqrt(4 × 5) km/hr v_res = 2sqrt(5) km/hr Step 3: Calculate the time taken to cross the river. The time taken is the river width divided by the effective velocity across the river: t = (d)/(v_res) t = 5 km2sqrt(5) km/hr To simplify, multiply the numerator and denominator by sqrt(5): t = 5sqrt(5)2sqrt(5)sqrt(5) hr t = 5sqrt(5)2 × 5 hr t = sqrt(5)2 hr As a decimal: t ≈ (2.236)/(2) hr t ≈ 1.118 hr The direction in which he must set off is 41.81^ upstream from the perpendicular to the river flow. The time it will take to cross the river is sqrt(5)2 hours. Send me the next one 📸