$A$ swimming pool is to be drained for cleaning. If $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain and $L=200(10-t)^{2}$,how fast is the water running out at the end of $5$ seconds? What is the average rate at which the water flows out during the first $5$ seconds?

(A) Given the volume of water $L$ at time $t$ is $L = 200(10-t)^2$.
The rate at which water flows out is given by $-\frac{dL}{dt}$.
First,find the derivative: $\frac{dL}{dt} = 200 \cdot 2(10-t) \cdot (-1) = -400(10-t)$.
Thus,the rate of flow is $-\frac{dL}{dt} = 400(10-t)$.
At $t = 5$ seconds,the rate is $400(10-5) = 400(5) = 2000 \text{ L/s}$.
To find the average rate during the first $5$ seconds,we calculate the total change in volume divided by the time interval: $\text{Average Rate} = \frac{L(0) - L(5)}{5 - 0}$.
$L(0) = 200(10-0)^2 = 200(100) = 20000 \text{ L}$.
$L(5) = 200(10-5)^2 = 200(25) = 5000 \text{ L}$.
$\text{Average Rate} = \frac{20000 - 5000}{5} = \frac{15000}{5} = 3000 \text{ L/s}$.