A rectangular swimming pool is 25 ft long and 20 1/2 ft wide, and it must be filled to a depth of 4 1/2 ft. One cubic foot of wa
1 answer:
ANSWER
307½ gallons of water
EXPLANATION
The volume of the rectangular swimming pool is
where l=25ft and w=20½ ft.
when it is filled to a depth of 4½ ft ,then the height is , h=4½ ft.
We substitute these values into the formula to get:
If one cubic foot of water is 7½ gallons, then 2306.25 cubic feet will be
We need 307.5 gallons of water to fill the pool.
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Answer:
a. the motion is positive in the time intervals:
The motion is negative in the time interval: (2,6)
b. S=7 m
c. distance=71m
Step-by-step explanation:
a. In order to solve part a. of this problem, we must start by determining when the velocity will be positive and when it will be negative. We can do so by setting the velocity equation equal to zero and then testing it for the possible intervals:
so let's solve this for t:
and now we factor it again:
(t-6)(t-2)=0
so we get the following answers:
t=6 and t=2
so now we can build our possible intervals:
and now we test each of the intervals on the given velocity equation, we do this by finding test values we can use to see how the velocity behaves in the whole interval:
[0,2) test value t=1
so:
v(1)=15 m/s
we got a positive value so the object moves in the positive direction.
(2,6) test value t=3
so:
v(3)=-9 m/s
we got a negative value so the object moves in the negative direction.
test value t=7
so:
v(1)=15 m/s
we got a positive value so the object moves in the positive direction.
the motion is positive in the time intervals:
The motion is negative in the time interval: (2,6)
b) in order to solve part b, we need to take the integral of the velocity function in the given interval, so we get:
so we get:
which simplifies to:
so now we evaluate the integral:
s=7 m
for part c, we need to evaluate the integral for each of the given intervals and add their magnitudes:
[0,2)
s(t)=\int\limits^2_0 {(3t^{2}-24t+36)} \, dt
so we get:
which simplifies to:
so now we evaluate the integral:
s=32 m
(2,6)
s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt
so we get:
which simplifies to:
so now we evaluate the integral:
s=-32 m
(6,7)
s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt
so we get:
which simplifies to:
so now we evaluate the integral:
s=7 m
and we now add all the magnitudes:
Distance=32+32+7=71m