1. The second Newton's Law
mv′=mg−31v Given m=2 kg,g=9.81 m/s2
v′+61v=9.81
v′=−61(v−58.86)
v−58.86dv=−61dt integrate
v(t)=58.86+c1e−t/6
v(0)=0=>c1=−58.86
v(t)=58.86−58.86e−t/6
v(1.5)=58.86−58.86e−1.5/6
v(1.5)=13.02 m/s
v(t)=−h′(t)
h(t)=−∫vdt=−∫(58.86−58.86e−t/6)dt
=−58.86t−353.16e−t/6+c2
h(0)=30=−353.16+c2=>c2=383.16
h(t)=−58.86t−353.16e−t/6+383.16
h(1.5)=−58.86(1.5)−353.16e−1.5/6+383.16
h(1.5)=19.83 m
2. Let s(t)= amount, in lb of salt at time t. Then we have
dtds=(rate of salt into tank) − (rate of salt out of tank)
dtds=2⋅12−50+(12−8)t8s So we get the differential equation
dtds=24−50+4t8s
dtds+25+2t4s=24 Integrating factor
μ(t)=e∫4dt/(25+2t)=(25+2t)2
dtd((25+2t)2s)=24(25+2t)2
∫d((25+2t)2s)=24∫(25+2t)2dt
(25+2t)2s=24(61)(25+2t)3+c1
s(t)=4(25+2t)+c1(25+2t)−2
s(0)=100+c1(25)−2=0
c1=−62500
s(t)=4(25+2t)−62500(25+2t)−2
s(5)=4(25+2(5))−62500(25+2(5))−2
s(5)=88.98 lbs(t)=4(25+2t)−62500(25+2t)−2=50 Let y(t)=s(t)−50
y(t)=4(25+2t)−62500(25+2t)−2−50 Solve graphically the equation y=0
t=2.47 min
Comments