Exercises 3.4⇒v(t)=9.5g√2(1−e−t/20).Sincex(0) = 0, integrating the above equation, we obtainx(t)=tZ0v(s)ds=tZ09.5g√2(1−e−s/20)ds=9.5g√2(s+20e−s/20)±±±s=ts=0.5g√2(te−t/20−20)≈131.8t+ 2636e−t/20−2636.The object reaches the end of the inclined plane whenx(t) = 131.8t+ 2636e−t/20−2636 = 10⇒t≈1.768 (sec).21.In this problem there are two forces acting on a sailboat: A constant horizontal force due tothe wind and a force due to the water resistance that acts in opposition to the motion of thesailboat. All of the motion occurs along a horizontal axis. On this axis, we choose the originto be the point where the hard blowing wind begins andx(t) denotes the distance the sailboat
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