Chapter 2wherevBis driver B’s constant velocity, andk>0 is a positive constant. Separating variableswe getdvAv2A=−kdt⇒ZdvAv2A=−Z⇒1vA(t)=kt+C.From the initial condition we fnd1vB=1vA(0)=k·0+C=C⇒C=1vB.ThusvA(t)=1+1/vB=vBvB.The ±unctions(t) there±ore satisfesdsdt=vBvB,s(0) = 0.Integrating we obtains(tZvBvBdt=1kln (vB+1)+C1.To fndC1we use the initial condition:0=s(0) =1kln (vBk·0+1)+C1=C1⇒C1=0.So,s(t1kln (vB+1).At the momentt=t1when driver A’s speed was halved, i.e.,
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.