A boy stands between two walls. He claps his hands and hears the first echo after 0.3 s and the second echo after 1.2 s. If the distance to the nearest wall is 50 m, calculate the speed of sound and t

The first echo is heard from the nearest wall. The sound travels to the wall and back to the boy.
The total distance covered by the sound for the first echo is $2d_1$.
Using the formula: distance = speed $\times$ time.

$$2d_1 = v \times t_1$$

Substitute the given values:

$$2 \times 50 \text{ m} = v \times 0.3 \text{ s}$$ $$100 \text{ m} = v \times 0.3 \text{ s}$$ $$v = \frac{100 \text{ m}}{0.3 \text{ s}}$$ $$v = \frac{1000}{3} \text{ m/s}$$ $$v \approx 333.33 \text{ m/s}$$

The speed of sound in air is $\boxed{\text{333.33 m/s}}$.

The second echo is heard from the farther wall. The sound travels to this wall and back to the boy.
The total distance covered by the sound for the second echo is $2d_2$.

$$2d_2 = v \times t_2$$

Substitute the calculated speed of sound and the given time for the second echo:

$$2d_2 = \left(\frac{1000}{3} \text{ m/s}\right) \times 1.2 \text{ s}$$ $$2d_2 = \frac{1000 \times 1.2}{3} \text{ m}$$ $$2d_2 = \frac{1200}{3} \text{ m}$$ $$2d_2 = 400 \text{ m}$$ $$d_2 = \frac{400 \text{ m}}{2}$$ $$d_2 = 200 \text{ m}$$

The boy is standing between the two walls. The total distance between the walls is the sum of the distance from the boy to the nearest wall ($d_1$) and the distance from the boy to the farther wall ($d_2$).

$$D = d_1 + d_2$$ $$D = 50 \text{ m} + 200 \text{ m}$$ $$D = 250 \text{ m}$$

The distance between the walls is $\boxed{\text{250 m}}$.