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Chapter 2: Q117P (page 39)

From January 26,1977, to September18,1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering30600km. In meters per second, what was the magnitude of his average velocity during that time period?

Short Answer

Expert verified

The average velocity of George Meegan of Great Britain is 0.15m/s,

Step by step solution

01

Given data

Total distance covered, 鈻�x=30600km

02

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:

vavg=鈻�x鈻�t ......(1)

Here, 鈻�xis the net displacement and 鈻�tis total time interval.

03

Determination of the average velocity of George Meegan

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

鈻�t=6years,7months,24days=2427days=2427脳24脳3600s=2.09脳108s

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

vavg=鈻�x鈻�t=30600km脳103m1km2.09脳108s鈮�0.15m/s

Thus, the average velocity of George Meegan of Great Britain islocalid="1654751953974" 0.15m/s.

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