In a flight of 600 Km an aircraft was slowed due to bad weather. Its average16 speed for the trip was reduced by 200 Km / hr and time of flight increased by 30 minutes. Find the original duration of flight.

Let's denote:

- \( d = 600 \, \text{km} \) (the distance of the flight),
- \( v \) as the original speed of the aircraft in \( \text{km/h} \),
- \( t \) as the original time of flight in hours.

### Step 1: Setting up the relationship between speed, distance, and time

We know that:
\[
v = \frac{d}{t} = \frac{600}{t}
\]
Now, due to bad weather, the speed was reduced by 200 km/h, so the new speed becomes \( v - 200 \). The time also increased by 30 minutes, or 0.5 hours, so the new time is \( t + 0.5 \).

Using the new speed and time, the distance can also be written as:
\[
v - 200 = \frac{600}{t + 0.5}
\]

### Step 2: Substituting \( v \) in terms of \( t \)

From the first equation, we have \( v = \frac{600}{t} \). Substituting this into the second equation gives:
\[
\frac{600}{t} - 200 = \frac{600}{t + 0.5}
\]

### Step 3: Solving the equation

We now solve the equation:
\[
\frac{600}{t} - 200 = \frac{600}{t + 0.5}
\]
Multiply both sides by \( t(t + 0.5) \) to eliminate the denominators:
\[
600(t + 0.5) - 200t(t + 0.5) = 600t
\]

Expanding both sides:
\[
600t + 300 - 200t^2 - 100t = 600t
\]
Simplify the equation:
\[
300 - 200t^2 - 100t = 0
\]
Divide the equation by 100 to make it simpler:
\[
3 - 2t^2 - t = 0
\]

### Step 4: Solving the quadratic equation

Rearrange the equation:
\[
2t^2 + t - 3 = 0
\]
Now, solve this quadratic equation using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = 1 \), and \( c = -3 \).

Substitute the values:
\[
t = \frac{-1 \pm \sqrt{1^2 - 4(2)(-3)}}{2(2)}
\]
\[
t = \frac{-1 \pm \sqrt{1 + 24}}{4}
\]
\[
t = \frac{-1 \pm \sqrt{25}}{4}
\]
\[
t = \frac{-1 \pm 5}{4}
\]

Thus, the two possible solutions are:
\[
t = \frac{-1 + 5}{4} = 1 \quad \text{(positive root)},
\]
\[
t = \frac{-1 - 5}{4} = -1.5 \quad \text{(negative root, discard)}.
\]

So, the original duration of the flight is \( t = 1 \) hour.

### Final Answer:
The original duration of the flight was **1 hour**.

---

Would you like any additional details or have any questions? Here are some related questions for further exploration:

1. What would happen if the speed reduction was different (e.g., 100 km/h)?
2. How would you calculate the total delay caused by bad weather?
3. What is the impact on fuel consumption due to a reduction in speed?
4. How would the solution change if the distance was 800 km instead of 600 km?
5. How does the quadratic equation help in solving real-world problems like this one?

**Tip:** When solving word problems, always define the variables clearly and set up equations based on relationships between them.

In a flight of 600 Km an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 Km / hr and time of flight increased by 30 minutes. Find the original duration of flight.

Explore a detailed solution to a time-speed-distance problem involving a 600 km flight. Learn how to calculate the original flight duration when speed is reduced by 200 km/h, and time increases by 30 minutes. This problem uses algebraic equations and the quadratic formula. Suitable for Grades 9-10.

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