Q117P - Fundamentals Of Physics

Question

From January $\mathbf{26}\mathbf{,}\mathbf{1977}$, to September$\mathbf{18}\mathbf{,}\mathbf{1983}\mathbf{ }$, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering $\mathbf{30600}\mathbf{ }\mathbf{km}\mathbf{ }$. In meters per second, what was the magnitude of his average velocity during that time period?

Step-by-Step Solution

Answer

The average velocity of George Meegan of Great Britain is  $0.15 \mathrm{m}/\mathrm{s}$,

Step 1: Given data

Total distance covered, $△\mathrm{x}=30600 \mathrm{km}$

Step 2: Understanding the average velocity

Average velocity may be defined as the ratio of net displacement to the total time interval. It is a vector quantity. 

The expression for the average velocity is given as follows:

 ${\mathrm{v}}_{\mathrm{avg}}=\frac{△\mathrm{x}}{△\mathrm{t}}$             ......(1)

Here, $△\mathrm{x}$ is the net displacement and $△\mathrm{t}$ is total time interval.

Step 3: Determination of the average velocity of George Meegan

The total time period from January 26, 1977 to September 18, 1983 is, 

$$\begin{gathered} △\mathrm{t} =6 \mathrm{years}, 7 \mathrm{months}, 24 \mathrm{days} \\ =2427 \mathrm{days} \\ =2427\times 24\times 3600 \mathrm{s} \\ =2.09\times {10}^8\mathrm{s} \end{gathered}$$

Using equation (i), the average velocity is calculated as follows: 

$$\begin{gathered} {\mathrm{v}}_{\mathrm{avg}}=\frac{△\mathrm{x}}{△\mathrm{t}} \\ =\frac{30600 \mathrm{km} \times {\displaystyle \frac{{10}^3\mathrm{m}}{1 \mathrm{km}}}}{2.09\times {10}^{8 }\mathrm{s}} \\ \approx 0.15 \mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the average velocity of George Meegan of Great Britain is$0.15 \mathrm{m}/\mathrm{s}$.