Explanation Given: Both sailboats have a mass of 60 kg . Both sailboats have an initial velocity of 2 m/sec . Constant force due to wind on both sailboats of 650 N Force due to water is proportional to the velocity of the boat. For the sailboat A the proportionality constants for the force due to water before the speed of 5 m/sec is reached and after are respectively, b 0 = 80 N-sec/m , b 1 = 60 N-sec/m . For the sailboat B the proportionality constants for the force due to water before the speed of 6 m/sec is reached and after are respectively, b 2 = 100 N-sec/m , b 3 = 50 N-sec/m . First leg of the race is 500 m long. Approach: Force due to wind is acting in direction of the velocity of the sailboat and force due to water resistance acting in direction opposite to velocity of the sailboat. For each sailboat derive the differential equation before and after planing. Solve for the distance travelled by each boat before planing and time taken to do this. Set time at which planing is achieved as t = 0 and solve for the time taken to travel the remaining distance in first leg. Calculation: We first solve for sailboat A The differential equation before planing is 60 d v d t = 650 − 80 v d v d t + 4 3 v = 65 6 Integrating factor for the above differential equation is: exp ( ∫ 4 3 d t ) = exp ( 4 t 3 ) Multiply by this integrating factor to get: d d t ( v exp ( 4 t 3 ) ) = 65 6 exp ( 4 t 3 ) Integrate to get: v exp ( 4 t 3 ) = 65 8 exp ( 4 t 3 ) + C v = 65 8 + C exp ( − 4 t 3 ) v ( 0 ) = 2 Apply the initial condition to get: v ( t ) = 65 8 − 49 8 exp ( − 4 t 3 ) Integrate again with respect to t to get: x ( t ) = 65 8 t + 147 32 exp ( − 4 t 3 ) + D Apply the initial condition: x ( 0 ) = 0 to get: x ( t ) = 65 8 t + 147 32 exp ( − 4 t 3 ) − 147 32 We compute the time when velocity becomes 5 m/sec by setting v ( t ) = 5 and solve for t . 5 = 65 8 − 49 8 exp ( − 4 t 3 ) 25 49 = exp ( − 4 t 3 ) t = − 3 4 ln ( 25 49 ) ≅ 0.505 sec The distance travelled by the boat A by this time is x ( 0.505 ) = 65 8 ⋅ 0.505 + 147 32 ( exp ( − 4 ⋅ 0.505 3 ) − 1 ) ≅ 1.85 m So amount of distance to be travelled by boat A after planing is 500 − 1.85 = 498.15 m We now solve for the motion after planing is achieved. The differential equation in this case is: 60 d v d t = 650 − 60 v d v d t + v = 65 6 Integrating factor for the above differential equation is: exp ( ∫ d t ) = exp ( t ) Multiply by this integrating factor to get: d d t ( v exp ( t ) ) = 65 6 exp ( t ) Integrate to get: v exp ( t ) = 65 6 exp ( t ) + E v = 65 6 + E exp ( − t ) v ( 0 ) = 5 Apply the initial condition to get: v ( t ) = 65 6 − 35 6 exp ( − t ) Integrate again with respect to t to get: x ( t ) = 65 6 t + 35 6 exp ( − t ) + F Apply the initial condition: x ( 0 ) = 0 to get: x ( t ) = 65 6 t + 35 6 exp ( − t ) − 35 6 Boat A needs to travel 498.15 m to finish first leg. We set: x = 498.15 to find time boat A takes to finish first leg

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 3.4, Problem 23E

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The sailboat which will be leading at the end of the first leg.

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Fundamentals of Differential Equations and Boundary Value Problems

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The sailboat which will be leading at the end of the first leg.

Solution Summary: The author explains that both sailboats have an initial velocity of 2m/sec and constant force due to water is proportional to the speed of the boat.

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To find: The sailboat which will be leading at the end of the first leg.

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Chapter 3 Solutions

To find:

The sailboat which will be leading at the end of the first leg.