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Chapter 2: Problem 117

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

Short Answer

Expert verified
George Meegan's average velocity was approximately 0.146 m/s.

Step by step solution

01

Calculate Total Time in Days

First, determine the total duration of the walk from January 26, 1977, to September 18, 1983. This period spans 6 years, 7 months, and 23 days. Calculate the total number of days, taking into account the days in each month and accounting for leap years where necessary. - **1977**: January 26 to December 31: 340 days - **1978**: 365 days - **1979**: 365 days - **1980** (leap year): 366 days - **1981**: 365 days - **1982**: 365 days - **1983**: January 1 to September 18: 261 days Total days = 340 + 365*5 + 366 + 261 = 2428 days.
02

Convert Days to Seconds

Next, convert the total duration from days to seconds to find the time interval in seconds.Since there are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, we have:\[\text{Total seconds} = 2428 \times 24 \times 60 \times 60 \]Calculating gives:\[\text{Total seconds} = 209,491,200 \text{ seconds}\]
03

Calculate Average Velocity in m/s

Now, use the formula for average velocity:\[\text{Average Velocity} = \frac{\text{Total Distance}}{\text{Total Time}}\]Given that the total distance is 30,600 km, convert this distance to meters by multiplying by 1,000:\[\text{Total Distance} = 30,600 \times 1,000 = 30,600,000 \text{ meters}\]Substitute into the formula:\[\text{Average Velocity} = \frac{30,600,000}{209,491,200} \approx 0.146 \text{ m/s}\]
04

Concluding Result

The calculation shows that George Meegan's average velocity over the entire journey was approximately 0.146 meters per second. This concludes the explanation of the average velocity calculation based on the given time period and distance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Conversion
Distance conversion is an essential concept when calculating average velocity. In the exercise, the total distance George Meegan walked was given as 30,600 kilometers. Since the standard unit for velocity in physics is meters per second, converting kilometers to meters is crucial. Remember that:
  • 1 kilometer equals 1,000 meters.
Therefore, to convert the distance from kilometers to meters, multiply the given distance by 1,000:
  • 30,600 kilometers 脳 1,000 = 30,600,000 meters.
This step ensures that the distance and time are in compatible units, which is vital for accurate velocity calculations.
Time Conversion
Time conversion is another critical step to find the average velocity accurately. In the context of the problem, the duration of George Meegan's journey was initially calculated over days. Here's how to convert those days into seconds, which is necessary for consistency in the velocity formula. First, note that:
  • 1 day has 24 hours,
  • 1 hour has 60 minutes,
  • 1 minute has 60 seconds.
Given that there are 2,428 days in the journey, the conversion formula is:
  • Total seconds = 2,428 days 脳 24 hours/day 脳 60 minutes/hour 脳 60 seconds/minute.
This results in a total time of 209,491,200 seconds, which is used in velocity calculations to ensure the units match those of the distance.
Velocity Formula
The velocity formula is a fundamental concept in physics. It provides a way to find the average velocity, which is defined as the total distance traveled divided by the total time taken. Mathematically, it is expressed as:
  • \( \text{Average Velocity} = \frac{\text{Total Distance}}{\text{Total Time}} \)
Applying it to George Meegan's journey:
  • Total Distance = 30,600,000 meters,
  • Total Time = 209,491,200 seconds,
  • Average Velocity = \( \frac{30,600,000}{209,491,200} \approx 0.146 \text{ m/s} \).
This average velocity of approximately 0.146 meters per second shows how far George traveled each second on average during his trek.
Leap Year Adjustment
When calculating time spans over several years, leap year adjustment is a necessary consideration. Leap years have an extra day (February 29), making a total of 366 days instead of the usual 365. In George Meegan's journey, one leap year needed to be accounted for:
  • 1977 - not a leap year,
  • 1978 - not a leap year,
  • 1979 - not a leap year,
  • 1980 - a leap year (366 days),
  • 1981 - not a leap year,
  • 1982 - not a leap year,
  • 1983 - counts up to September 18.
By including the leap year in 1980, adjustments ensure the accuracy of the total time calculated, ensuring precise results for average velocity. Missed leap years can lead to significant discrepancies over long periods.

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Most popular questions from this chapter

To stop a car, first you require a certain reaction time to begin braking; then the car slows at a constant rate. Suppose that the total distance moved by your car during these two phases is \(56.7 \mathrm{~m}\) when its initial speed is \(80.5 \mathrm{~km} / \mathrm{h}\), and \(24.4 \mathrm{~m}\) when its initial speed is \(48.3 \mathrm{~km} / \mathrm{h}\). What are (a) your reaction time and (b) the magnitude of the acceleration?

A rock is thrown vertically upward from ground level at time \(t=0 .\) At \(t=1.5 \mathrm{~s}\) it passes the top of a tall tower, and \(1.0 \mathrm{~s}\) later it reaches its maximum height. What is the height of the tower?

A ball of moist clay falls \(15.0 \mathrm{~m}\) to the ground. It is in contact with the ground for \(20.0 \mathrm{~ms}\) before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?

The position of a particle moving along an \(x\) axis is given by \(x=12 t^{2}-2 t^{3}\), where \(x\) is in meters and \(t\) is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at \(t=3.0 \mathrm{~s}\). (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and \((\mathrm{g})\) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at \(t=0\) )? (i) Determine the average velocity of the particle between \(t=0\) and \(t=3 \mathrm{~s}\).

In 1889 , at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope \(3.7 \mathrm{~m}\). In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?
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