The main objective of this question is to find the time that elapses for the boat to arrive at the wave’s trough.
This question uses the concept of crest, trough, and wavelength of the wave. A surface wave’s crest is a region where the medium’s displacement is greatest. The smallest or minimum level in a cycle is called a trough since it is the opposite of a crest, while the wavelength of a wave signal traveling through space along a wire is the separation between two corresponding points in the adjoining cycles.
Expert Answer
We have to find the time that elapses for the boat to arrive at the wave’s trough.
The wave wavelength is:
\[\lambda \space = \space 100m \]
The wave speed is:
\[v \space = \space 55 \space k \frac{m}{h}\]
We know that:
\[d \space = \space \frac{\lambda}{2} \]
By putting the values, we get:
\[= \space \frac{160}{2} \]
\[= \space 80 m \]
As:
\[v \space = \space \frac{d}{t} \]
And time $ t $ is:
\[t \space = \space \frac{d}{v} \]
By putting the values, we get:
\[ \space = \space \frac{80}{55} \space \times \space \frac{1}{1000} \space \times \space \frac{3600}{1} \]
\[ \space = \space \frac{80}{55} \space \times \space \frac{1}{1000} \space \times \space 3600 \]
\[ \space = \space \frac{80}{55} \space \times \space 10^-3 \space \times \space 3600 \]
\[ \space = \space 1.4545 \space \times \space 10^-3 \space \times \space 3600 \]
\[ \space = \space 5236.3636 \space \times \space 10^-3 \]
\[ \space = \space 5.23 \space s \]
Thus, the time calculated is $ 5.23 \space s $.
Numerical Answer
The time elapsed is $ 5.23 \space s $.
Example
A storm is generating waves that are hitting a motionless boat in the ocean. The waves’ wavelength is $ 180 m $, and their speed is $ 55 km/h $. The boat is near a wave’s peak. How long does it take for the boat to arrive at the wave’s trough?
We have to find the time that elapses for the boat to arrive at the wave’s trough.
The wave wavelength is given as:
\[\lambda \space = \space 100m \]
The wave speed is equal to:
\[v \space = \space 55 \space k \frac{m}{h}\]
We know that:
\[d \space = \space \frac{\lambda}{2} \]
By putting the values, we get:
\[ \space= \space \frac{180}{2} \]
\[ \space = \space 90 m \]
As we know:
\[v \space = \space \frac{d}{t} \]
And time $ t $ is:
\[t \space = \space \frac{d}{v} \]
By putting the values, we get:
\[ \space = \space \frac{90}{55} \space \times \space \frac{1}{1000} \space \times \space \frac{3600}{1} \]
\[ \space = \space \frac{90}{55} \space \times \space \frac{1}{1000} \space \times \space 3600 \]
\[ \space = \space \frac{90}{55} \space \times \space 10^-3 \space \times \space 3600 \]
\[ \space = \space 1.6363 \space \times \space 10^-3 \space \times \space 3600 \]
\[ \space = \space 5890.9091 \space \times \space 10^-3 \]
\[ \space = \space 5.89 \space s \]
Thus, the time elapsed is $ 5.89 \space s $.