An aeroplane is stopped for half an hour and now it has to cover a distance of 1500 km in given time so its speed is increased by 250 km/h find its initial speed.

Answer

To find the initial speed of the aeroplane, we can follow these steps:

### Step 1: Define Variables
Let:
- \( S_1 \) = initial speed of the aeroplane (in km/h)
- \( S_2 \) = final speed of the aeroplane after increasing speed (in km/h)

According to the problem, we know that:
\[ S_2 = S_1 + 250 \]

### Step 2: Understand the Time Difference
The aeroplane was stopped for half an hour (0.5 hours). After the stop, it needs to cover a distance of 1500 km. The time difference caused by the speed increase is equal to the time saved from the stop, which is 0.5 hours.

### Step 3: Use the Distance Formula
The distance covered can be expressed in terms of speed and time:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]

For the two speeds, we can express the time taken to cover the distance at each speed:
- Time taken at initial speed \( S_1 \):
\[ t_1 = \frac{1500}{S_1} \]
- Time taken at increased speed \( S_2 \):
\[ t_2 = \frac{1500}{S_2} \]

### Step 4: Set Up the Equation
The time difference between the two speeds is equal to the time saved from the stop:
\[ t_1 - t_2 = 0.5 \]
Substituting the expressions for \( t_1 \) and \( t_2 \):
\[ \frac{1500}{S_1} - \frac{1500}{S_2} = 0.5 \]

### Step 5: Substitute \( S_2 \) in the Equation
Substituting \( S_2 = S_1 + 250 \) into the equation:
\[ \frac{1500}{S_1} - \frac{1500}{S_1 + 250} = 0.5 \]

### Step 6: Solve the Equation
To eliminate the fractions, we can multiply through by \( S_1(S_1 + 250) \):
\[ 1500(S_1 + 250) - 1500S_1 = 0.5S_1(S_1 + 250) \]

This simplifies to:
\[ 1500 \times 250 = 0.5S_1(S_1 + 250) \]
\[ 375000 = 0.5S_1(S_1 + 250) \]
Multiplying both sides by 2:
\[ 750000 = S_1(S_1 + 250) \]

### Step 7: Rearrange the Equation
Rearranging gives us:
\[ S_1^2 + 250S_1 - 750000 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( S_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 1, b = 250, c = -750000 \):
\[ S_1 = \frac{-250 \pm \sqrt{250^2 - 4 \times 1 \times (-750000)}}{2 \times 1} \]
\[ S_1 = \frac{-250 \pm \sqrt{62500 + 3000000}}{2} \]
\[ S_1 = \frac{-250 \pm \sqrt{3062500}}{2} \]
\[ S_1 = \frac{-250 \pm 1750}{2} \]

Calculating the two potential solutions:
1. \( S_1 = \frac{1500}{2} = 750 \) (valid)
2. \( S_1 = \frac{-2000}{2} = -1000 \) (not valid)

### Conclusion
Thus, the initial speed of the aeroplane is:
\[ S_1 = 750 \text{ km/h} \]

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