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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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
We have given
where c is integrating constant,to be determined with initiatial condition.
t=0 , R=15000
15000=500 x 0+ 10 x 0+c
Water in dam at any time t is
Water in dam after 10 hrs.
R=500 x 10+ 10 x100 +15000
Ans. There will be 21000 L water in the dam after 10 houurs .
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