The speed of sound in air is 60 m/s. How far from the centre of a storm is an observer who hears a thunder clap 4s after the flash of the lightning?

Light travels so much faster than sound that the flash of lightning reaches the observer effectively at the instant it is produced. The 4 s delay is therefore the whole time the sound took to cover the distance from the storm to the observer, and the distance follows from the definition of speed: \[d = v \times t.\]

Substituting the values given in the question: \[d = 60 \times 4 = 240\ \text{m}.\] The observer is 240 m from the centre of the storm.

Two points are worth fixing. First, use the speed value the question supplies, not the familiar figure for air at room temperature; this question deliberately sets the speed at \(60\ \text{m s}^{-1}\), so answering 340 m by using \(340\ \text{m s}^{-1}\) and a time of 1 s, or by multiplying the wrong pair of numbers, ignores the given data. Second, do not halve the time as you would in an echo calculation. An echo travels to a reflector and back, so there the distance is \(\frac{vt}{2}\); thunder makes a one-way trip, so the full time is used. Exam reminder: decide first whether the sound path is one-way or a there-and-back journey before applying the speed equation.