Answer:
a. -900 L/min
b. Vo = 63,000 Liters
c. V ( t ) = 63,000 - 900*t
d. 7,200 Liters
Explanation:
Solution:-
We have a swimming pool which is drained at a constant linear rate. Certain readings were taken for the volume of water remaining in the pool ( V ) at different instances time ( t ) as follows.
t = 20 mins , V = 45,000 Liters
t = 70 mins , V = 0 Liters
To determine the rate ( m ) at which water drains from the pool. We can use the linear rate formulation as follows:
![m = (V_2 - V_1)/(t_2 - t_1) \\\\m = (0 - 45000)/(70 - 20) \\\\m = (-45,000)/(50) = - 900 (L)/(min)](https://img.qammunity.org/2021/formulas/mathematics/high-school/878spbbgrln3fudkx83dm3he2u99whpxku.png)
We can form a linear relationship between the volume ( V ) remaining in the swimming pool at time ( t ), using slope-intercept form of linear equation as follows:
![V = m*t + V_o](https://img.qammunity.org/2021/formulas/mathematics/high-school/hc2ns0o9wubkv373bwt0jz8fiqrrqicams.png)
Where,
m: the rate at which water drains
Vo: the initial volume in the pool at time t = 0.
We can use either of the data point given to determine the initial amount of volume ( Vo ) in the pool.
![0 = -900*( 70 ) + V_o\\\\V_o = 63,000 L](https://img.qammunity.org/2021/formulas/mathematics/high-school/19r3d67aquelkks5i025id02k7z3ntd33i.png)
We can now completely express the relationship between the amount of volume ( V ) remaining in the pool at time ( t ) as follows:
![V ( t ) = 63,000 - 900*t](https://img.qammunity.org/2021/formulas/mathematics/high-school/ki5gad2loxby6f4u276gwbjrpxg6wi13k8.png)
We can use the above relation to determine the amount of volume left after t = 62 minutes as follows:
![V ( 62 ) = 63,000 - 900*(62 )\\\\V ( 62 ) = 63,000 - 55,800\\\\V ( 62 ) = 7,200 L](https://img.qammunity.org/2021/formulas/mathematics/high-school/j8tzsb3eoxgklyokniok26qymhjrho6vnd.png)