Asked by Harry
The rate of flow of water into a dam is given by R=500+20t L h^-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: 21 000L
Answer: 21 000L
Answers
Answered by
Bosnian
If R' = 500 +20 t
then:
Rate of flow of water into the dam:
dR / dt = 500 + 20 t
dR = 500 dt + 20 t ∙ dt
Integrating both sides:
R = 500 t + 20 t² / 2 + C
R = 10 t² + 500 t + C
C = integration constant
Initial condition:
t = 0 , R = 15 000
R = 10 t² + 500 t + C
15 000 = 10 ∙ 0² + 500 ∙ 0 + C
15 000 = C
C = 15 000
R = 10 t² + 500 t + C
R = 10 t² + 500 t + 15 000
After 10 hours:
R = 10 ∙ 10² + 500 ∙ 10 + 15 000
R = 10 ∙ 100 + 5 000 + 15 000
R = 1 000 + 5 000 + 15 000
R = 21 000 L
then:
Rate of flow of water into the dam:
dR / dt = 500 + 20 t
dR = 500 dt + 20 t ∙ dt
Integrating both sides:
R = 500 t + 20 t² / 2 + C
R = 10 t² + 500 t + C
C = integration constant
Initial condition:
t = 0 , R = 15 000
R = 10 t² + 500 t + C
15 000 = 10 ∙ 0² + 500 ∙ 0 + C
15 000 = C
C = 15 000
R = 10 t² + 500 t + C
R = 10 t² + 500 t + 15 000
After 10 hours:
R = 10 ∙ 10² + 500 ∙ 10 + 15 000
R = 10 ∙ 100 + 5 000 + 15 000
R = 1 000 + 5 000 + 15 000
R = 21 000 L
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.