Question

The summer has ended and it’s time to drain the swimming pool. 20 minutes after pulling the plug, there is still 45 000L of water in the pool. The pool is empty after 70 minutes. Calculate the rate that the water is draining out of the pool. (Hint: remember this line is sloping down to the right) b) Calculate how much water was in the pool initially (at time 0). I think it was 80 000 c) Write an equation for this relationship. d) Use your equation to calculate how much water is in the pool at 62 minutes.

Answer 1

Step-by-step explanation:

At t = 20 min, V = 45,000 L.

At t = 70 min, V = 0 L.

a) The rate is the slope:

m = ΔV/Δt

m = (0 L − 45,000 L) / (70 min − 20 min)

m = -900 L/min

b) Using the slope to find V when t = 0 min:

m = ΔV/Δt

-900 L/min = (V − 0 L) / (0 min − 70 min)

V = 63,000 L

c) Use slope-intercept form to find the equation.

V = -900t + 63,000

d) At t = 62 min:

V = -900(62) + 63,000

V = 7200

Answer 2

Answer:

a. -900 L/min

b. Vo = 63,000 Liters

c. V ( t ) = 63,000 - 900*t

d. 7,200 Liters

Step-by-step 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:

                             $m = \frac{V_2 - V_1}{t_2 - t_1} \\\\m = \frac{0 - 45000}{70 - 20} \\\\m = \frac{-45,000}{50} = - 900 \frac{L}{min}$

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:

                              $V = m*t + V_o$

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.

                           $0 = -900*( 70 ) + V_o\\\\V_o = 63,000 L$

We can now completely express the relationship between the amount of volume ( V ) remaining in the pool at time ( t ) as follows:

                             $V ( t ) = 63,000 - 900*t$

We can use the above relation to determine the amount of volume left after t = 62 minutes as follows:

                           $V ( 62 ) = 63,000 - 900*(62 )\\\\V ( 62 ) = 63,000 - 55,800\\\\V ( 62 ) = 7,200 L$