a swimming pool is 50 m long and 20 m wide. its depth decreases linearly along the length from 9 m to 1 m…

Explanation:

Step1: Find the volume of water after 250 min

The inflow rate is $1m^{3}/min$. So the volume of water $V$ after $t = 250$ min is $V=1\times250 = 250m^{3}$.

Step2: Set up the volume - height relationship

Let the length of the pool be $L = 50m$, width be $w=20m$. Assume the water - level height is $h$. The cross - sectional area of the water in the pool at height $h$ is a trapezoid. The depth of the pool varies linearly from $1m$ to $9m$ over a length of $50m$. The average depth of the water at height $h$ is related to the volume. The volume of water $V$ in the pool is $V = A\times w$, where $A$ is the cross - sectional area of the water. The cross - sectional area of the water forms a trapezoid. The relationship between the volume $V$ and the height $h$ of the water is $V=\left(\frac{(1 + 1+\frac{8h}{50})}{2}\right)\times50\times20$. Simplify it: $V=(2+\frac{8h}{50})\times50\times10=(100 + 8h)\times10 = 1000+80h$.

Step3: Differentiate the volume equation with respect to time

Differentiate $V = 1000+80h$ with respect to time $t$. We get $\frac{dV}{dt}=80\frac{dh}{dt}$. We know that $\frac{dV}{dt}=1m^{3}/min$.

Step4: Solve for $\frac{dh}{dt}$

Substitute $\frac{dV}{dt}=1$ into $\frac{dV}{dt}=80\frac{dh}{dt}$. Then $1 = 80\frac{dh}{dt}$, so $\frac{dh}{dt}=\frac{1}{80}m/min$.

Step5: Find the volume of the pool

The volume of the pool $V_{pool}=\frac{(1 + 9)}{2}\times50\times20=5000m^{3}$.

Step6: Find the time to fill the pool

Since the inflow rate is $1m^{3}/min$, the time $T$ to fill the pool is $T=\frac{V_{pool}}{\frac{dV}{dt}}=\frac{5000}{1}=5000$ min.

Answer:

The water is rising at a rate of $\frac{1}{80}m/min$ 250 min after the filling begins. It will take 5000 min to fill the pool.