dv/dt = -2 + 1/10 t
v = -2t + 1/20 t^2 + C
since dv/dt = 0 when t=20, that means that
v(20) = 500, so
-2*20 + 1/20 * 20^2 + C = 500
C = 27
Thus, v(t) = 1/20 t^2 - 2t + 27
(D) ∫[0,20] v(t) dt = 620/3
(E) with the tap fully on, at 2 m^3/s, it would take (300 - 620/3)/2 = 140/3 seconds to drain the extra 280/3 m^3
Only part (d) and (e) please:)
A tap on a large tank is gradually turned off so as not to create any hydraulic shock. As a consequence, the flow rate while the tap is being turned off is given by dV/dt = −2 + 1/10t m3/s.
A) What is the initial flow rate when the tap is fully on?
B) How long does it take to turn the tap off?
C) Given that when the tap has been turned off there are still 500 m3 of water left in the tank, find V as a function of t.
D) Hence find how much water is released during the time it takes to turn the tap off.
E) Suppose that it is necessary to let out a total of 300 m3 from the tank. How long should the tap be left fully on before gradually turning it off?
2 answers
The answer for D) is apparently 20m^3 released
and for E) is 140 seconds= 2 mins and 20 seconds
Thank you:)
and for E) is 140 seconds= 2 mins and 20 seconds
Thank you:)