A loaded boat weighs 981kg.The force exerted upon the boat in the direct of motion by the oars,is equivalent to a constant force of 15N and the resistance in Newton to the motion is equal to twice the speed of the boat . If th boat starts from rest, determine;
(i)the limiting speed
(ii)speed at the end of 10 sec.
i)
ma=15−2vma=15-2vma=15−2v
mdvdt=15−2vm\frac{dv}{dt}=15-2vmdtdv=15−2v
981⋅dvdt=15−2v981\cdot\frac{dv}{dt}=15-2v981⋅dtdv=15−2v
dt=98115−2vdvdt=\frac{981}{15-2v}dvdt=15−2v981dv
t=∫98115−2vdv=−9812ln(15−2v)+ct=\int \frac{981}{15-2v}dv=-\frac{981}{2}ln(15-2v)+ct=∫15−2v981dv=−2981ln(15−2v)+c
v=0v=0v=0 at t=0t=0t=0
c=9812ln15c=\frac{981}{2}ln15c=2981ln15
t=−9812ln(15−2v)+9812ln15t=-\frac{981}{2}ln(15-2v)+\frac{981}{2}ln15t=−2981ln(15−2v)+2981ln15
e2t/981=1515−2ve^{2t/981}=\frac{15}{15-2v}e2t/981=15−2v15
v(t)=7.5−7.5e2t/981v(t)=7.5-\frac{7.5}{e^{2t/981}}v(t)=7.5−e2t/9817.5
ii)
v(10)=7.5−7.5e2⋅10/981=0.15v(10)=7.5-\frac{7.5}{e^{2\cdot 10/981}}=0.15v(10)=7.5−e2⋅10/9817.5=0.15 m/s
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