Q2E - University Physics with Modern Physics

Question

In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 km away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin at the nest and extend the +x-axis to the release point, what was the bird’s average velocity in  (a) for the return flight and (b) for the whole episode, from leaving the nest to returning?

Step-by-Step Solution

Answer

(a)The average velocity of seabird in return flight is,  $v_{avg}^{return}=4.42 m/s$

(b) The average velocity of seabird for the whole trip is, $v_{avg}^{trip}=\text{0m/s}$

Step 1: Identification of the given data

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

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

  $$\begin{gathered} 1 \text{km}=1000 \text{m} \\ \text{5150} \text{km}=\text{5150}\times \text{1000 m}=5150000 \text{m} \end{gathered}$$

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

 Since,

  $$\begin{gathered} \text{1day}=\text{24} \text{hrs} \\ \text{1hr}=\text{3600} \text{s } \end{gathered}$$

Therefore,

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

Step 2: Calculation of the Average velocity of the seabird in return flight

The average velocity of the seabird can be expressed as,

$v_{avg}=\frac{d}{t}$ ……………………….(i)

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 $4.42 m/s$

Step 3: Calculation of the Average velocity of the seabird for the whole trip

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  $0m/s$.