Please help with some Calculus to the Phy World questions :) (1 Viewer)

[s8891]

Hi

I'm having problems with some of these problems. It would be great if you could post up worked solutions for them

1. A rectangular vessel is divided into two equal compartments by a vertical porous membrane. Liquid in one compartment, initially at a depth of 20cm, passes into the other compartment, initially empty, at a rate proportional to the difference in levels.
a) If the depth of liquid in one of the vessels at any time t minutes is x cm, show that dx/dt=k(20-2x)
b) Show that x=10[1-e^(-2kt)]
c) If the level in the second compartment rises 2cm in the first 5 minutes, after what time will the difference in levels be 2cm?

2. A body falls from rest in a medium and its velocity, v metres/second, after t seconds is given by: v=80[1-e^(-0.4t)
a) Show that the acceleration is proportional to 80-v
b) Calculate the distance fallen in the first 5 seconds.
c) Calculate the distance fallen when v=60.

Thanks in advance. Will be repping

 

[zeebobDD]

2. Acc. is d(v)/dt

a= 80(-e^-0.4t * (-0.4)) = 0.4 * 80e^-0.4t but V=80-80e^-0.4t

therefore a = 0.4 * (80-V)

b) x - intgeral v dt where v=80(1-e^-0.4t)

dx = 80(1-e^-0.4t) dt

integrate 80(1-e^-0.4t) dt from 5 to 0 since its asking for the first 5 seconds
and you end up with 227m

c) V *(dv/dx) = (80-V) * 0.4

therefore V dv / (80-V) = 0.4 dx

integrating LHS from 60 to 0 as q' asks for v=60 seconds

then just integrate and you will arrive with the answer

 

[s8891]

2. Acc. is d(v)/dt

a= 80(-e^-0.4t * (-0.4)) = 0.4 * 80e^-0.4t but V=80-80e^-0.4t

therefore a = 0.4 * (80-V)

b) x - intgeral v dt where v=80(1-e^-0.4t)

dx = 80(1-e^-0.4t) dt

integrate 80(1-e^-0.4t) dt from 5 to 0 since its asking for the first 5 seconds
and you end up with 227m

c) V *(dv/dx) = (80-V) * 0.4

therefore V dv / (80-V) = 0.4 dx

integrating LHS from 60 to 0 as q' asks for v=60 seconds

then just integrate and you will arrive with the answer

can you show the working for part c? im not getting the right answer :S

 

[zeebobDD]

so 0.4x = int from 60to 0 v dv/ (80-v)

for the integrand you can either use long division or manipulate the function

so 0.4x = (-1 + 80/(80-v)) dv

0.4x = -v-[1/80(Ln(80-v)] dv

sub back in the limits 60 to 0

and you will end up with 80Ln(4)-60 but 0.4x = 80Ln(4)-60 so x = 127.25 approx.

 

[Timske]

can you show the working for part c? im not getting the right answer :S

<a href="http://www.codecogs.com/eqnedit.php?latex=Sub ~ v = 60 \\ e^{-0.4t} = \frac{-v@plus;80}{80} \\ -0.4t = \ln(\frac{2}{8}) \\ t= 3.466 ~(3~dp) \\Sub~ t~in~ x=80t @plus; 200e^{-0.4t} - 200 \\ x = 127m" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Sub ~ v = 60 \\ e^{-0.4t} = \frac{-v+80}{80} \\ -0.4t = \ln(\frac{2}{8}) \\ t= 3.466 ~(3~dp) \\Sub~ t~in~ x=80t + 200e^{-0.4t} - 200 \\ x = 127m" title="Sub ~ v = 60 \\ e^{-0.4t} = \frac{-v+80}{80} \\ -0.4t = \ln(\frac{2}{8}) \\ t= 3.466 ~(3~dp) \\Sub~ t~in~ x=80t + 200e^{-0.4t} - 200 \\ x = 127m" /></a>

 

[Sanjeet]

 

[karnbmx]

Like a boss.