91Ó°ÊÓ

The distance at which the seabird is released =$5150\text{km}$

The value of distance in km can be converted in m using,

role="math" localid="1655367436412" $$\begin{gathered} 1  \text{km}=1000 \text{m} \\ \text{5150} \text{km}=\text{5150}×\text{1000 m}=5150000\text{m} \end{gathered}$$

Time taken by the seabird to return to its nest = 13.5 days

Since,

role="math" localid="1655367527326" $$\begin{gathered} \text{1day}=\text{24} \text{hrs} \\ \text{1hr}=\text{3600} \text{s} \end{gathered}$$

$$\begin{gathered} \text{13.5} \text{days}=\text{13.5}\text{days}×\left(\frac{\text{24}\text{hrs}}{1\text{day}}\right)×\left(\frac{3600\text{s}}{1\text{hr}}\right) \\ =1166400\text{s} \end{gathered}$$

The average velocity of the seabird can be expressed as,

$v_{avg}=\frac{d}{t}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(¾±)

Where, d and t are the displacement and time, respectively.

Substituting the given values in the equation (i) for the return flight,

$\begin{array}{rcl}{v}_{avg}^{return}& =& \frac{5150000m}{1166400s}\\ & =& 4.42m/s\end{array}$

Thus, the average velocity in the return flight is$4.42m/s$

As, the seabird has returned to the same place from where she has started, the total displacement of the bird is zero. Therefore, substituting the values in equation (i) will give the average velocity during trip,

$\begin{array}{rcl}{v}_{avg}^{trip}& =& \frac{0m}{ts}\\ & =& 0m/s\end{array}$

Thus, the average velocity for the whole trip is $0m/s$.