#### Answer

The skipper first finds the frequency from the observed period.
$f = \frac{1}{T} = \frac{1}{5 s}= 0.2 Hz$.
The distance between wave crests is the wavelength. Finally, the wave speed is frequency multiplied by wavelength.
$v = f \lambda = (0.2 Hz)(15 m) = 3 \frac{m}{s}$.
Alternatively, 15 meters of wave passes in 5 seconds. The speed of the wave must be $\frac{15 m}{5 s} = 3 \frac{m}{s}$.