A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/h from the usual speed. Find its usual speed.

Options

A. 720 km/h
B. 740 km/h
C. 730 km/h
D. 750 km/h

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Define the Variables**:
Let the usual speed of the plane be \( x \) km/h.

2. **Calculate the Usual Time**:
The distance to the destination is 1500 km. The time taken at the usual speed can be calculated using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1500}{x} \text{ hours} \]

3. **Calculate the Increased Speed**:
Since the plane left 30 minutes late, it had to increase its speed by 250 km/h to reach on time. Therefore, the new speed is:
\[ x + 250 \text{ km/h} \]

4. **Calculate the Actual Time with Increased Speed**:
The time taken to cover the same distance at the increased speed is:
\[ \text{Actual Time} = \frac{1500}{x + 250} \text{ hours} \]

5. **Set Up the Equation**:
The difference in time between the usual time and the actual time is 30 minutes, which is \( \frac{1}{2} \) hours. Thus, we can set up the equation:
\[ \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \]

6. **Clear the Fractions**:
To eliminate the fractions, multiply through by \( 2x(x + 250) \):
\[ 2x(x + 250) \left( \frac{1500}{x} - \frac{1500}{x + 250} \right) = 2x(x + 250) \cdot \frac{1}{2} \]
This simplifies to:
\[ 3000(x + 250) - 3000x = x(x + 250) \]

7. **Simplify the Equation**:
Simplifying gives:
\[ 3000 \cdot 250 = x^2 + 250x \]
\[ 750000 = x^2 + 250x \]

8. **Rearrange to Form a Quadratic Equation**:
Rearranging gives:
\[ x^2 + 250x - 750000 = 0 \]

9. **Factor the Quadratic Equation**:
We can factor this quadratic equation. We need two numbers that multiply to \(-750000\) and add to \(250\). The numbers are \(1000\) and \(-750\):
\[ (x + 1000)(x - 750) = 0 \]

10. **Solve for \( x \)**:
Setting each factor to zero gives:
\[ x + 1000 = 0 \quad \text{or} \quad x - 750 = 0 \]
Thus, \( x = -1000 \) (not valid since speed cannot be negative) or \( x = 750 \).

11. **Conclusion**:
Therefore, the usual speed of the plane is:
\[ \boxed{750 \text{ km/h}} \]