Explanation:

- Let the usual speed of the airplane be \( x \) km/h.

- The usual time to cover 2100 km is \( \frac{2100}{x} \) hours.

- The plane took off 45 minutes late, or \( \frac{3}{4} \) hour late.

- So, it needs to cover 2100 km in \( \frac{2100}{x} - \frac{3}{4} \) hours to be on time.

- The increased speed is 40% more, so the new speed is \( x + 0.4x = 1.4x \).

- Using the formula for time, set up the equation:

$$

\frac{2100}{1.4x} = \frac{2100}{x} - \frac{3}{4}

$$

- Solve for x:

$$

1.4x \cdot \left(\frac{2100}{1.4x}\right) = 1.4x \left(\frac{2100}{x} - \frac{3}{4}\right)

$$

- Cross-multiply and solve:

$$

2100 = 2940 - \frac{3}{4} \times 1.4x \rightarrow \frac{3}{4} \times 1.4x = 2940 - 2100 = 840

$$

- Solve for x:

$$

x = \frac{840 \times 4}{3 \times 1.4} = 800 \text{ km/h}

$$

- Increased speed is:

$$

1.4 \times 800 = 1120 \text{ km/h} \quad \textcolor{green}{\checkmark}

$$