An anti-tank gun fires a shell at a tank. The shell hits the…

An anti-tank gun fires a shell at a tank. The shell hits the tank $t_1 = 0.6\ \mathrm{s}$ after firing, and the sound of the explosion is heard at the gun $t_2 = 2.1\ \mathrm{s}$ after firing. What is the distance from the gun to the tank, and what is the average speed of the shell? The speed of sound is $c = 340\ \mathrm{m/s}$.

Explanation of Method

Finding a distance from the delay of a sound.

Step 1: Split the 2.1 seconds

The 2.1 s contain two trips over the same distance $d$: the shell flying to the tank, and then the sound of the explosion coming back: \[t_{\text{sound}} = t_2 - t_1 = 2.1\ \mathrm{s} - 0.6\ \mathrm{s} = 1.5\ \mathrm{s}.\]

$x\ (\mathrm{m})$ $t\ (\mathrm{s})$ $510$ $0$ $0$ $0.6$ $2.1$ $\text{explosion at the tank}$ $\text{shell}$ $\text{sound, } 340\ \mathrm{m/s}$ $t_1 = 0.6\ \mathrm{s}$ $t_2 - t_1 = 1.5\ \mathrm{s}$

shell (gun → tank)
sound (tank → gun)

Position–time diagram. The shell goes out to the tank in 0.6 s; the sound of the explosion then needs another 1.5 s to come back over the same 510 m.

Step 2: Distance to the tank

\[ d = c\,t_{\text{sound}} = 340\ \mathrm{m/s} \cdot 1.5\ \mathrm{s}. \]

\[d = 510\ \mathrm{m}\]

Step 3: Average speed of the shell

\[ v = \frac{d}{t_1} = \frac{510\ \mathrm{m}}{0.6\ \mathrm{s}}. \]

\[v = 850\ \mathrm{m/s}\]

Check

$ \dfrac{510}{850}\ \mathrm{s} + \dfrac{510}{340}\ \mathrm{s} = 0.6\ \mathrm{s} + 1.5\ \mathrm{s} = 2.1\ \mathrm{s}\ \checkmark $

Result

\[\boxed{\,d = 510\ \mathrm{m}\,}\]

\[\boxed{\,v = 850\ \mathrm{m/s}\,}\]

Remark

Note. The shell is 2.5 times faster than sound, so the way back (sound) takes 2.5 times longer than the way out (shell): $1.5 = 2.5 \cdot 0.6$.