Two boats A and B move away from a buoy along mutually perpendicular lines and, after moving equal distances, return. Each boat's speed relative to water is 1.2 times the stream speed. The ratio of their total times tₐ/tᵦ is:
Options

Question

Two boats A and B move away from a buoy along mutually perpendicular lines and, after moving equal distances, return. Each boat's speed relative to water is 1.2 times the stream speed. The ratio of their total times tₐ/tᵦ is:

Options

G1 Physics · Accepted Answer

Let stream speed be u and boat speed relative to water be v=1.2u. Suppose each boat goes a distance l from the buoy and returns. Boat A moves along the river: tₐ = l/(v+u) + l/(v-u) = 2lv/(v²-u²). Boat B must move along a line perpendicular to the river. Its effective speed across the river is √(v²-u²). Hence tᵦ = 2l/√(v²-u²). Therefore tₐ/tᵦ = v/√(v²-u²). With v=1.2u: tₐ/tᵦ =1.2/√(1.44-1) =1.2/√0.44 =6/√11.