Question

solving using the speed of sound
if a storm is 7.5 kilometers away, how much time is expected between observations of lightning and thunder? round your answer to one decimal place.
21.9 × seconds

Explanation:

Step1: Recall the speed of sound in air

The speed of sound in air is approximately \( 343 \, \text{m/s} \), but for the approximation related to thunder and lightning, a common rule is that sound travels about \( 0.34 \, \text{km/s} \) (or we can use the conversion: since \( 1 \, \text{km} = 1000 \, \text{m} \), and speed of sound \( v = 343 \, \text{m/s}=\frac{343}{1000} \, \text{km/s}= 0.343 \, \text{km/s} \)). The time \( t \) is given by the formula \( t=\frac{d}{v} \), where \( d \) is the distance and \( v \) is the speed.

Step2: Identify distance and speed

The distance \( d = 7.5 \, \text{km} \), and we use the speed of sound \( v = 0.343 \, \text{km/s} \) (or the approximate \( 0.34 \, \text{km/s} \), but let's use the more accurate \( 343 \, \text{m/s} \) by converting distance to meters: \( d = 7.5\times1000 = 7500 \, \text{m} \), \( v = 343 \, \text{m/s} \))

Step3: Calculate time

Using \( t=\frac{d}{v} \), substitute \( d = 7500 \, \text{m} \) and \( v = 343 \, \text{m/s} \)
\( t=\frac{7500}{343}\approx21.9 \) (Wait, no, wait: Wait, actually, the common approximation is that for every kilometer, the time delay is about 3 seconds (since speed of sound is about \( \frac{1}{3} \, \text{km/s} \), so \( t = \frac{d}{v} \), if \( v=\frac{1}{3} \, \text{km/s} \), then \( t = 3d \) seconds per kilometer. Wait, let's check:

Wait, speed of sound in air at 20°C is about 343 m/s. So to find time for distance \( d \) (in meters), \( t=\frac{d}{343} \) seconds.

Given \( d = 7.5 \, \text{km}=7500 \, \text{m} \)

So \( t=\frac{7500}{343}\approx21.9 \) seconds? Wait, but let's recalculate:

\( 343\times21 = 7203 \), \( 7500 - 7203 = 297 \), \( 297\div343\approx0.866 \), so total \( t\approx21.866\approx21.9 \) seconds. Wait, but maybe the approximation is using \( v = 330 \, \text{m/s} \) (speed of sound in air at 0°C). Let's check with \( v = 330 \, \text{m/s} \):

\( t=\frac{7500}{330}\approx22.7 \) seconds. But the initial answer was 21.9, which is with \( v = 343 \, \text{m/s} \). Wait, maybe the problem uses the approximation that sound travels 1 km in about 2.9 seconds (since \( 1000/343\approx2.915 \)). So for 7.5 km, \( 7.5\times2.915\approx21.86\approx21.9 \) seconds. So the calculation is correct.

Wait, but let's do the calculation properly:

\( t=\frac{d}{v} \), where \( d = 7.5 \, \text{km} = 7500 \, \text{m} \), \( v = 343 \, \text{m/s} \)

\( t=\frac{7500}{343}\approx21.9 \) seconds (rounded to one decimal place). Wait, but maybe the user made a mistake, but according to the calculation, 7500 divided by 343 is approximately 21.9. Wait, let's check with a calculator: 34321.9 = 34320 + 343*1.9 = 6860 + 651.7 = 7511.7, which is a bit more than 7500, so 21.9 is a bit high. Let's do 7500/343:

34321 = 7203, 7500 - 7203 = 297, 297/343 ≈ 0.866, so 21.866, which rounds to 21.9. So the answer is 21.9 seconds. Wait, but maybe the problem expects using the approximation that time (in seconds) is distance (in kilometers) multiplied by 3 (since 1 km takes about 3 seconds). Then 7.53 = 22.5 seconds. But that's a rough approximation. However, the more accurate calculation with 343 m/s gives ~21.9. So the initial answer of 21.9 is correct? Wait, maybe the system marked it wrong by mistake, but according to the calculation, 7500/343 ≈21.9.

Answer:

21.9 seconds