3 boats set out to cross a river that has a current. The speedof the 2 boats rel

Question

3 boats set out to cross a river that has a current. The speedof the 2 boats relative to the water is the same. BoatA points at an angle of 30 degrees upsteam from a pointdirectly across a river. Boat B points directlyacross and Boat C 20 degrees downsteam from apoint directly across the river.
Which takes the last time, most time, and moderate time? 3 boats set out to cross a river that has a current. The speedof the 2 boats relative to the water is the same. BoatA points at an angle of 30 degrees upsteam from a pointdirectly across a river. Boat B points directlyacross and Boat C 20 degrees downsteam from apoint directly across the river.
Which takes the last time, most time, and moderate time?

Explanation / Answer

let the time taken by boat A,B and C to cross the river bet1,t2 and t3 respectively therefore time taken by boat 1 is t1 = (d/V1y + Vw) = (d/V1 * sin + Vw) d is the width of the river,V1 is the speed of boat A, =30o and Vw is the speed of current in the river or t1 = (d/V1 * sin(30o) + Vw) = (d/0.5 * V1 + Vw)------------(1) time taken by boat B is t2 = (d/V2) -------------(2) V2 is the speed of boat B time taken by boat C is t3 = (d/V3y - Vw) = (d/V3 * sin1 - Vw) V3 is the speed of boat C and 1 = 20o or t3 = (d/V3 * sin(20o) - Vw) = (d/0.342 * V3 -Vw) --------------(3) let the speed of boat A and boat C be same that is V1 = V3therefore from (1) and (3) we get t1 = (d/0.5 * V1 + Vw) -------------(4) and t3 = (d/0.342 * V1 - Vw) --------------(5) (t1/t3) = [(d/0.5 * V1 + Vw)/(d/0.342 * V1 - Vw)] or (t1/t3) = (0.342 * V1 - Vw/0.5 * V1 + Vw) = [0.342 -(Vw/V1)/0.5 + (Vw/V1)] when V1 >> Vw we get (t1/t3) = (0.342/0.5) = 0.684 or t1 = 0.684 * t3 -------------(6) the boat B takes the smallest time to cross the river as ittravels directly across the river from equation (6) we can conclude that the time taken by boatA is more when compared to boat C therefore we get t1 > t3 > t2 or t3 = (d/V3 * sin(20o) - Vw) = (d/0.342 * V3 -Vw) --------------(3) let the speed of boat A and boat C be same that is V1 = V3therefore from (1) and (3) we get t1 = (d/0.5 * V1 + Vw) -------------(4) and t3 = (d/0.342 * V1 - Vw) --------------(5) (t1/t3) = [(d/0.5 * V1 + Vw)/(d/0.342 * V1 - Vw)] or (t1/t3) = (0.342 * V1 - Vw/0.5 * V1 + Vw) = [0.342 -(Vw/V1)/0.5 + (Vw/V1)] when V1 >> Vw we get (t1/t3) = (0.342/0.5) = 0.684 or t1 = 0.684 * t3 -------------(6) the boat B takes the smallest time to cross the river as ittravels directly across the river from equation (6) we can conclude that the time taken by boatA is more when compared to boat C therefore we get t1 > t3 > t2

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