A road runs midway between two parallel rows of buildings. A motorist moving with a speed of 36 Km/h sounds the horn. He hears the echo one second after he has sounded the horn: Then the distance between the two rows of buildings is. (Velocity of sound in air is 330 m/s)

Options

A. `80 sqrt17m`
B. `40 sqrt17 m`
C. `30 sqrt10 m`
D. `330m`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the speed of the motorist from km/h to m/s
The speed of the motorist is given as 36 km/h. To convert this to meters per second (m/s), we use the conversion factor:

\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]

Calculating this gives:

\[ \text{Speed} = 36 \times \frac{5}{18} = 10 \, \text{m/s} \]

### Step 2: Calculate the distance traveled by the motorist in 1 second
Since the motorist is moving at a speed of 10 m/s and he hears the echo after 1 second, the distance he travels in that time is:

\[ \text{Distance (PQ)} = \text{Speed} \times \text{Time} = 10 \, \text{m/s} \times 1 \, \text{s} = 10 \, \text{m} \]

### Step 3: Determine the distance the sound travels
The sound travels to the buildings and back to the motorist. Since he hears the echo after 1 second, the total distance traveled by the sound is:

\[ \text{Total distance (sound)} = \text{Speed of sound} \times \text{Time} = 330 \, \text{m/s} \times 1 \, \text{s} = 330 \, \text{m} \]

### Step 4: Calculate the distance from the point where the horn was sounded to one building
Since the sound travels to one building and back, the distance from the horn to one building (let's call it x) is half of the total distance traveled by the sound:

\[ x = \frac{\text{Total distance (sound)}}{2} = \frac{330 \, \text{m}}{2} = 165 \, \text{m} \]

### Step 5: Use the Pythagorean theorem to find the distance between the two rows of buildings
Let y be the perpendicular distance from the midpoint of the road to the building. We can use the Pythagorean theorem:

\[ y^2 + \left(\frac{10}{2}\right)^2 = x^2 \]

Substituting the values we have:

\[ y^2 + 5^2 = 165^2 \]

Calculating \(165^2\) and \(5^2\):

\[ y^2 + 25 = 27225 \]

Now, solving for y:

\[ y^2 = 27225 - 25 = 27200 \]

Taking the square root:

\[ y = \sqrt{27200} \approx 165 \, \text{m} \]

### Step 6: Calculate the total distance between the two rows of buildings
The total distance (D) between the two rows of buildings is:

\[ D = 2y = 2 \times 165 \approx 330 \, \text{m} \]

### Final Answer
The distance between the two rows of buildings is approximately **330 meters**.

---