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 **s****mallest 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 $.