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?

Short answer

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

Step-by-step solution

Given data

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

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.

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 islocalid="1654751953974" $0.15\mathrm{m}/\mathrm{s}$.