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November 8th, 2010, 07:57 AM   #1
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physicis problem.. Sand dunes on a desert island move as san

1 ) Sand dunes on a desert island move as sand is swept up the windward side to settle in the leeward side. Such “walking” dunes have been known to travel 20 feet in a year and can travel as much as 100 feet per year in particularly windy times. Calculate the average speed in each case in m/s.

(b) Fingernails grow at the rate of drifting continents, about 10 mm/yr. Approximately how long did it take for North America to separate from Europe, a distance of about 3 000 mi?

[color=#0080FF]a )

20ft = 32180m

average speed = 32180/100 = 321.8


t = 3000/10 = 300[/color]


2 ) A bristlecone pine tree has been know to take 4000 years to grow to a height of 20 ft.
a)Find the avearge speed of growth in m/s
b) In contrast the faster growing palnt is the gaint kelp which can grow at rate of 2 feet in one day

[color=#0040FF]a )
4000 years conver to secound

now 20 ft to m = 6.1
avearge speed = 6.1/345.6X10^-6 = 1.7650

b) I don't understand this ? [/color]


3 ) Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 70 mi/h. (a) Assuming that they start at the same point, how much sooner does the faster car arrive at a destination 10 mi away? (b) How far must the faster car travel before it has a 15-min lead on the slower car?
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November 8th, 2010, 10:37 AM   #2
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Math Focus: Calculus/ODEs
Re: physicis problem.. Sand dunes on a desert island move as

1. There are 0.3048 meters per foot. There are approximately 31556925.9936 seconds in a year.



(b) We can use , where time is t, average speed is and displacement is x. We are given x = 3000 miles, and:



(a) The slower car takes to travel 10 miles, whereas the faster car takes . To find how much sooner the faster car arrives than the slower car, we find the difference in their times.

(b) 15 minutes is .25 hour. So we can use:

=\frac{\frac{1}{4} \text{ hr.}}{\left(\frac{1}{55}-\frac{1}{70}\right)\text{ \frac{hr.}{mi}}}=\frac{385}{6}\text{ mi.}=64\frac{1}{6}\text{ mi.}" />

Another way to find this is to use the result from part a to set up the following ratios:

Cross multiply:

Multiply through by :

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