Understanding the Aircraft Flight Problem

This problem involves calculating the original flight duration of an aircraft given the total distance, a reduction in average speed due to bad weather, and the corresponding increase in flight time.

Here's a breakdown of the information given:

• Total Distance: 900 km
• Speed Reduction: 300 km/h
• Time Increase: 30 minutes (which is equal to 0.5 hours)

Setting Up the Equations

Let's define the variables:

• Let the original speed of the aircraft be s km/h.
• Let the original time of the flight be t hours.

We know the relationship between distance, speed, and time is: Distance = Speed × Time.

Using the original conditions, we can write the first equation:

$$ 900 = s \times t $$ (Equation 1)

Due to bad weather, the conditions changed:

• The new average speed is $ s' = s - 300 $ km/h.
• The new flight time is $ t' = t + 0.5 $ hours.

Using the new conditions and the same distance, we can write the second equation:

$$ 900 = s' \times t' $$

Substituting the expressions for $ s' $ and $ t' $:

$$ 900 = (s - 300)(t + 0.5) $$ (Equation 2)

Solving for the Original Flight Time

Now we need to solve these two equations to find the value of t.

1. From Equation 1, we can express the original speed s in terms of t:

   $$ s = \frac{900}{t} $$

2. Substitute this expression for s into Equation 2:

   $$ 900 = \left(\frac{900}{t} - 300\right)(t + 0.5) $$

3. Expand the right side of the equation:

   $$ 900 = \frac{900}{t} \times t + \frac{900}{t} \times 0.5 - 300 \times t - 300 \times 0.5 $$

   $$ 900 = 900 + \frac{450}{t} - 300t - 150 $$

4. Simplify the equation by subtracting 900 from both sides and combining constants:

   $$ 0 = \frac{450}{t} - 300t - 150 $$

   $$ 150 = \frac{450}{t} - 300t $$

5. To eliminate the fraction, multiply the entire equation by t (assuming $t \neq 0$):

   $$ 150t = 450 - 300t^2 $$

6. Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

   $$ 300t^2 + 150t - 450 = 0 $$

7. Simplify the quadratic equation by dividing all terms by 150:

   $$ 2t^2 + t - 3 = 0 $$

8. Solve the quadratic equation. We can solve this by factoring:

   We look for two numbers that multiply to ($2 \times -3 = -6$) and add to ($1$). These numbers are $3$ and $-2$.

   $$ 2t^2 + 3t - 2t - 3 = 0 $$

   Factor by grouping:

   $$ t(2t + 3) - 1(2t + 3) = 0 $$

   $$ (t - 1)(2t + 3) = 0 $$

9. This gives two possible values for t:
   • $ t - 1 = 0 \implies t = 1 $
   • $ 2t + 3 = 0 \implies t = -\frac{3}{2} $
10. Since the time of flight cannot be negative, we discard the solution $ t = -\frac{3}{2} $.

Therefore, the original duration of the flight is 1 hour.

Verification

Let's check if this result is correct:

• Original time ($t$) = 1 h
• Original speed ($s$) = $ \frac{900}{1} = 900 $ km/h
• New time ($t'$) = $ 1 + 0.5 = 1.5 $ h
• New speed ($s'$) = $ 900 - 300 = 600 $ km/h
• Check distance with new values: $ s' \times t' = 600 \times 1.5 = 900 $ km.

The calculated distance matches the given distance, confirming our answer.