Exercises 3.4Solving this linear equation and using the initial condition,v(A)1(0) = 2, we getv(A)1(t)=658−498e−4t/3,and sox(A)1(ttZ0±658−498e−4s/3²ds=658t−14732(e−4t/3−1).The boat A will have the velocity 5 m/sec att=t∗satisfying658−498e−4t∗/3=5⇒t∗=−3ln(25/49)4≈0.5 (sec),and it will bex(A)1(t∗658t∗−14732(e−4t∗/3−1)≈1.85 (m)away from the starting point or, equivalently, 500−1.85 = 498.15 meters away from the Fnish.Similarly to (3.12), resetting the time, we obtain an equation of the motion of the boat Astarting from the moment when its velocity reaches 5 m/sec. Denoting byx(A)2(t) the distancepassed by the boat A and byv(A)2(t) its velocity, we getx(A)2(0) = 0,v(A)2(0) = 5, andmdv(A)2dt= 650−b2v(A)2⇒dv(A)2dt=656−v(A)2⇒v(A)2(t656−356e−t⇒x(A)2(ttZ0±656−356e−s²ds=656t+356(e−t−1).Solving the equationx(A)2(t) = 498.15, we Fnd the time (counting from the moment when the
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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 Berkeley.