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Question

We are filling a swimming pool with water. At Noon on Sunday the water was 51 inches deep at 3 pm, its depth was 6 feet. 1. What was the pools depth at 1:00pm on Sunday? 2. What was the pools depth at 6pm sunday evening? 3. How many hours after noon on sunday was the water two yards deep? 4. We stopped filling the pool when the depth of the water reached eight feet. How many hours did we spend filling the pool after noon on Sunday? When did we start filling the pool? To write the expression , define a variable for the time after noon that the pool was filled , and use the variable to write a rule for the total depth in inches of the pool.

Answer 1

Diamond,

 

First we find how much water was added in the 3 hrs between noon and 3 pm

At 3 pm d₃ = 6 ft = 72 in and at noon d_n = 51 in, then in 3 hrs the depth rose by (72 - 51) in = 21 in, so the rate at which the water is rising is 7 in/hr.

 

we can write an expression for the depth as a function of time, with t = 0 at noon:

 

d = 7·t + 51, where d = depth in inches, and t = time in hours.

 

For 1 pm t = 1 so d₁ = 7 + 51 = 58 in

 

For 6 pm t = 6 so d₆ = 7·6 + 51 = 42 + 51 = 93 in

 

To find the time when the water was 2 yards deep, first change depth to inches = 72 in, and then solve for t, but we can save ourselves some work by noting that 2 yards = 6 ft, which was our depth at 3 pm.

 

So t = 3 pm is when the water was 2 yds deep.

 

Solve for t again to find the time when the water reached a depth of 8 ft = 96 in.

 

7·t = d - 51 ⇒ t = (d - 51)/7 = (96 - 51)/7 = 45/7 = 6.43 hrs  (This is not the total time spent filling the pool, this is the time after noon that it took to go from 51 in deep to 8 ft deep)

 

To find the total time spent filling the pool we need to know the time started.   To reach a depth of 51 in took time

t = (51 in(/(7 in/hr) = 7.29 hrs = 7½ hr.

 

Subtract 7½ hrs from noon (12:00) gives 4:30 am the time starting to fill the pool.

 

Add 7.5 + 6.43 ≈ 13.9 hr =13 hr 56 min (total time spent filling the pool)