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The rate of flow of water into a dam is given by R'= 500+20t Lh^-1. If there is 15,000L...

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bubbletrea | Student, Grade 11 | Honors

Posted May 27, 2013 at 6:21 AM via web

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The rate of flow of water into a dam is given by R'= 500+20t Lh^-1. If there is 15,000L of water initially in the dam, how much water will there be in the dam after 10 hours?

 

answer is 21,000L

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science2014 | Valedictorian

Posted June 30, 2014 at 7:23 PM (Answer #3)

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Rate of flow of water into the dam, R'= 500+20t = dR/dt

or dR = (500 + 20t) dt

Integrating both sides:

R = 500t + 20 * (t^2/2) + c ,   c being integration constant

Initial condition: t = 0, R = 15,000

So  15,000 = 500*0 + 20*(0^2/2) + c

or c = 15,000

Plugging in the value of c = 15,000

the amount of water going into the dam at time t is given by:

R = 500t + 20 * (t^2/2) + 15,000

So after 10 hours,

R = 500*10 + 20 * (10^2/2) + 15,000

= 5,000 + 1,000 + 15,000

= 21,000 L

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pramodpandey | College Teacher | Valedictorian

Posted May 27, 2013 at 7:08 AM (Answer #2)

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We have given

`(dR)/(dt)=500+20t`

`dR=(500+20t)dt`

on integration

`R=500t+20t^2/2+c`

`R=500t+10t^2+c`

where c is integrating constant,to be determined with initiatial condition.

t=0   , R=15000

15000=500 x 0+ 10 x 0+c

c=15000 L

Water in dam at any time t is

R=500t+10t^2+15000

Water in dam after 10 hrs.

R=500 x 10+ 10 x100 +15000

R=5000+1000+15000

R=21000 L

Ans.  There will be  21000 L water  in the dam after 10 houurs .

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