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In Problem 21 it is observed that when the velocity of the sailboat reaches 5 m/sec, the boat begins to rise out of the water and 鈥減lane.鈥 When this happens, the proportionality constant for the water resistance drops to b0 = 60 N-sec/m. Now find the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind as it is planning?

Short Answer

Expert verified
  • The limiting velocity of the sailboat is 10 m/second.
  • The equation of motion is x(t)=10t+256(e-1.2t-1).

Step by step solution

01

Find the equation of velocity

There are two forces are acting in the direction of the boat and in the opposite direction of the boat respectively.

F=600NF2=60v

Now

role="math" localid="1664211510034" mdvdt=600-60vmdvdt=600-60vdvdt=12-1.2vdv12-1.2v=dt(Integratingonbothsides)-11.2ln12-1.2v=t+Cln12-1.2v=-1.2t-1.2C12-1.2v=k.e-1.2t

Put v = 5, t = 0 then k = 6

12-1.2v=6.e-1.2tv(t)=10-5.e-1.2t

02

Find the value of equation of motion

x(t)=0tv(s)dsx(t)=0t10-5e-1.2sds=10s+5e-1.2s1.20tx(t)=10t+256(e-1.2t-1)

Hence,The equation of motion isx(t)=10t+256(e-1.2t-1).

03

Find the limiting velocity

The limiting velocity of the sailboat is

limtv(t)=limt10-5-1.2t=10

Hence, the Limiting velocity is 10m/sec.

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