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Chapter 1: Problem 18

The approximate maximum speeds of various animals follow, but in different units of speed. Convert these data to \(\mathrm{m} / \mathrm{s}\), and thereby arrange the animals in order of increasing maximum speed: squirrel, \(19 \mathrm{~km} / \mathrm{h}\); rabbit, 30 knots; snail, \(0.030 \mathrm{mi} / \mathrm{h}\); spider, \(1.8 \mathrm{ft} / \mathrm{s}\); cheetah, \(1.9 \mathrm{~km} / \mathrm{min}\); human, \(1000 \mathrm{~cm} / \mathrm{s} ;\) fox, \(1100 \mathrm{~m} / \mathrm{min} ;\) lion, \(1900 \mathrm{~km} /\) day.

Short Answer

Expert verified
After converting all the given speeds to \( \mathrm{m} / \mathrm{s} \) and arranging them in increasing order, the result is: Snail (0.0134 \( \mathrm{m} / \mathrm{s} \)), Spider (0.55 \( \mathrm{m} / \mathrm{s} \)), Squirrel (5.28 \( \mathrm{m} / \mathrm{s} \)), Human (10 \( \mathrm{m} / \mathrm{s} \)), Rabbit (15.42 \( \mathrm{m} / \mathrm{s} \)), Fox (18.33 \( \mathrm{m} / \mathrm{s} \)), Lion (22.01 \( \mathrm{m} / \mathrm{s} \))

Step by step solution

01

Calculate Squirrel's Speed

First, let's convert squirrel's speed from \(\mathrm{km} / \mathrm{h}\) to \(\mathrm{m} / \mathrm{s}\). The conversion factor between these units is \( \frac{1 \mathrm{km}}{1000 \mathrm{m}} \) and \( \frac{1 \mathrm{h}}{3600 \mathrm{s}} \). Hence, the speed of squirrel is \( \frac{19 \mathrm{km} / \mathrm{h}}{1} \cdot \frac{1000 \mathrm{m}}{1 \mathrm{km}} \cdot \frac{1 \mathrm{h}}{3600 \mathrm{s}} = 5.28 \mathrm{m} / \mathrm{s} \)
02

Calculate Rabbit's Speed

Secondly, the rabbit's speed needs to be converted from knots to \( \mathrm{m} / \mathrm{s} \). One knot equals \(0.514 \mathrm{m} / \mathrm{s}\). So, the speed of a rabbit is \(30 \mathrm{knots} \cdot 0.514 \mathrm{m} / \mathrm{s} = 15.42 \mathrm{m} / \mathrm{s} \)
03

Calculate Snail's Speed

Thirdly, convert the snail's speed from \(\mathrm{mi} / \mathrm{h}\) to \(\mathrm{m} / \mathrm{s}\). One \(\mathrm{mi} / \mathrm{h}\) is approximately \(0.447 \mathrm{m} / \mathrm{s}\). Therefore, the snail's speed is \(0.03 \mathrm{mi} / \mathrm{h} \cdot 0.447 \mathrm{m} / \mathrm{s} = 0.0134 \mathrm{m} / \mathrm{s} \)
04

Calculate Spider's Speed

4th, convert spider's speed from \(\mathrm{ft} / \mathrm{s}\) to \(\mathrm{m} / \mathrm{s}\). The conversion factor is \( \frac{1 \mathrm{ft}}{0.305 \mathrm{m}} \), therefore, the speed of spider is \(1.8 \mathrm{ft} / \mathrm{s} \cdot \frac{0.305 \mathrm{m}}{1 \mathrm{ft}} = 0.55 \mathrm{m} / \mathrm{s}\)
05

Calculate Human's Speed

The 5th step consists of converting the human speed from \(\mathrm{cm} / \mathrm{s}\) to \(\mathrm{m} / \mathrm{s}\). In this case, it's needed to divide by 100, hence the human speed is \( \frac{1000 \mathrm{cm} / \mathrm{s}}{100} = 10 \mathrm{m} / \mathrm{s} \)
06

Calculate Fox's Speed

Using the conversion factor \( \frac{1 \mathrm{min}}{60 \mathrm{s}} \), the fox speed is \( \frac{1100 \mathrm{m} / \mathrm{min}}{1} \cdot \frac{1 \mathrm{min}}{60 \mathrm{s}} = 18.33 \mathrm{m} / \mathrm{s} \)
07

Calculate Lion's Speed

Finally, to find out lion's speed, you should use the conversion factors \( \frac{1 \mathrm{km}}{1000 \mathrm{m}} \), \( \frac{1 \mathrm{day}}{24 \mathrm{h}} \) and \( \frac{1 \mathrm{h}}{3600 \mathrm{s}} \), hence the lion's speed is \( \frac{1900 \mathrm{km} / \mathrm{day}}{1} \cdot \frac{1000 \mathrm{m}}{1 \mathrm{km}} \cdot \frac{1 \mathrm{day}}{24 \mathrm{h}} \cdot \frac{1 \mathrm{h}}{3600 \mathrm{s}} = 22.01 \mathrm{m} / \mathrm{s} \)

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Most popular questions from this chapter

Assuming that the length of the day uniformly increases by \(0.001 \mathrm{~s}\) in a century, calculate the cumulative effect on the measure of time over 20 centuries. Such a slowing down of the Earth's rotation is indicated by observations of the occurrences of solar eclipses during this period.

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