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Outflow: \( \frac{V(t)}{200} \times 0.1 \times 5 \, \text{L/min} = \frac{0.5V(t)}{200} \).

Let \( S(t) \) be the salt in kg at time \( t \).

The differential equation is \( \frac{dS}{dt} = 0.5 - \frac{0.5S}{200} \).

Solving gives \( S(t) = 10e^{-0.0025t} + 20(1 - e^{-0.0025t}) \).

At \( t = 30 \), \( S(30) = 10e^{-0.075} + 20(1 - e^{-0.075}) \approx 14.31 \) kg.

#### 14.31

A car accelerates uniformly from rest to 60 m/s in 10 seconds, then maintains constant speed for 20 seconds, and finally decelerates uniformly to rest in 5 seconds. What is the total distance traveled?

Acceleration phase: \( a = \frac{60}{10} = 6 \, \text{m/s}^2 \); distance \( = \frac{1}{2} \times 6 \times 10^2 = 300 \, \text{m} \).

Constant speed: \( d = 60 \times 20 = 1200 \, \text{m} \).

Deceleration phase: \( a = \frac{-60}{5} = -12 \, \text{m/s}^2 \); distance \( = 60 \times 5 + \frac{1}{2}(-12)(5^2) = 300 - 150 = 150 \, \text{m} \).

florida lottery result

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