Question #267082
  1. An object has a mass of 2kg is dropped from the top of a building 30 meters tall. the initial velocity is zero. As it falls, the object encounters air resistance that is equal to 1/3 v. what is the velocity and attitude of the object after 1.5 seconds? Ans. 13.02m/s; 19.83m
  2. A tank initially contains 50 gal of fresh water. At t=0, brine solution containing 2 lb of salt per gallon is poured into the tank at the rate of 12 gal/min, while the well stirred mixture leaves the tank at the rate of 8 gal/min. what is the amount of salt at the end of 5 minutes? How long will it take to obtain an amount of 50 lb? ans. 88098 lb; 2.47 min
1
Expert's answer
2021-11-17T15:39:54-0500

1. The second Newton's Law


mv=mg13vmv'=mg-\dfrac{1}{3}v

Given m=2 kg,g=9.81 m/s2m=2\ kg, g=9.81\ m/s^2


v+16v=9.81v'+\dfrac{1}{6}v=9.81

v=16(v58.86)v'=-\dfrac{1}{6}(v-58.86)

dvv58.86=16dt\dfrac{dv}{v-58.86}=-\dfrac{1}{6}dt

integrate


v(t)=58.86+c1et/6v(t)=58.86+c_1e^{-t/6}

v(0)=0=>c1=58.86v(0)=0=>c_1=-58.86

v(t)=58.8658.86et/6v(t)=58.86-58.86e^{-t/6}

v(1.5)=58.8658.86e1.5/6v(1.5)=58.86-58.86e^{-1.5/6}

v(1.5)=13.02 m/sv(1.5)=13.02\ m/s

v(t)=h(t)v(t)=-h'(t)

h(t)=vdt=(58.8658.86et/6)dth(t)=-\int v dt=-\int(58.86-58.86e^{-t/6})dt

=58.86t353.16et/6+c2=-58.86t-353.16e^{-t/6}+c_2

h(0)=30=353.16+c2=>c2=383.16h(0)=30=-353.16+c_2=>c_2=383.16

h(t)=58.86t353.16et/6+383.16h(t)=-58.86t-353.16e^{-t/6}+383.16

h(1.5)=58.86(1.5)353.16e1.5/6+383.16h(1.5)=-58.86(1.5)-353.16e^{-1.5/6}+383.16

h(1.5)=19.83 mh(1.5)=19.83\ m

2. Let s(t)=s(t) = amount, in lb of salt at time t.t. Then we have

dsdt=\dfrac{ds}{dt}=(rate of salt into tank) − (rate of salt out of tank)


dsdt=2128s50+(128)t\dfrac{ds}{dt}=2\cdot12-\dfrac{8s}{50+(12-8)t}

So we get the differential equation


dsdt=248s50+4t\dfrac{ds}{dt}=24-\dfrac{8s}{50+4t}

dsdt+4s25+2t=24\dfrac{ds}{dt}+\dfrac{4s}{25+2t}=24

Integrating factor


μ(t)=e4dt/(25+2t)=(25+2t)2\mu(t)=e^{\int 4dt/(25+2t)}=(25+2t)^{2}

ddt((25+2t)2s)=24(25+2t)2\dfrac{d}{dt}((25+2t)^{2}s)=24(25+2t)^{2}

d((25+2t)2s)=24(25+2t)2dt\int d((25+2t)^{2}s)=24\int(25+2t)^{2}dt

(25+2t)2s=24(16)(25+2t)3+c1(25+2t)^{2}s=24(\dfrac{1}{6})(25+2t)^{3}+c_1

s(t)=4(25+2t)+c1(25+2t)2s(t)=4(25+2t)+c_1(25+2t)^{-2}

s(0)=100+c1(25)2=0s(0)=100+c_1(25)^{-2}=0

c1=62500c_1=-62500

s(t)=4(25+2t)62500(25+2t)2s(t)=4(25+2t)-62500(25+2t)^{-2}

s(5)=4(25+2(5))62500(25+2(5))2s(5)=4(25+2(5))-62500(25+2(5))^{-2}

s(5)=88.98 lbs(5)=88.98\ lbs(t)=4(25+2t)62500(25+2t)2=50s(t)=4(25+2t)-62500(25+2t)^{-2}=50

Let y(t)=s(t)50y(t)=s(t)-50

y(t)=4(25+2t)62500(25+2t)250y(t)=4(25+2t)-62500(25+2t)^{-2}-50

Solve graphically the equation y=0y=0

t=2.47 mint=2.47\ min



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