Thunder and Lightning

Short Solution: Since the answer options are rounded to the nearest kilometre, we realise that the answer is not a very close approximation. We make an assumption from this:

Assumption: The light has travelled instantaneously (which, of course, it hasn't really).

Therefore, we have assumed that the lightning struck when Tom saw it: this shows us that the lightning is $340 \times 6=2040$ meters away (using the old $\text {Distance}=\text{Speed} \times \text{Time}$ ), which is roughly $\boxed{2 \text{km}}$ .

Sound travels 340m/s * 6s = 2040m or about 2km (the speed of light is so fast it doesn't make a difference over such a small distance).

We know, if the speed is V and time need to travel is t, then distance covered is, D = V * t. Here the light and sound covers the same distance.

Let, light needs X seconds to travel, so sound needs (X + 6) seconds. As the light and sound covers same distance, so the equation will be like this: 3 10^8 X = 340*(X+6), and you can get the value of X from this. After getting this value, place X's value in any side of the equation stated above, and you will get the answer. Just consider that you may need to round the answer to nearest kilometre.

C= d/t, where c= speed of sound (as he hears the thunder), d= distance from the thunder and t= time taken to hear the thunder.

Rearranging, d= cxt. So d = 340 m/s x 6s= 2040 m/ 2 km.

d= average velocity * time elapsed

As speed of light is too fast to be considered in this situation, you just have to estimate the speed of sound that could roughly round to 1/3 km/sec in the air at ground level so 6 seconds will place you about 2 km from the lightning. Note: To get more precision you would need air atmospheric pressure, humidity, temperature etc, to estimate the density in what the sound travel.