My Math Forum physicis problem.. Sand dunes on a desert island move as san

 Physics Physics Forum

 November 8th, 2010, 07:57 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 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 b) 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 =345.6X10^-6 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?
 November 8th, 2010, 10:37 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 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. (a1) $|\bar{v}|=\frac{20}{1}\text{ \frac{ft.}{yr.}}\cdot \frac{0.3048\text{ m.}}{1\text{ ft.}}\cdot \frac{1\text{ yr.}}{31556925.9936\text{ s.}}\approx 1.932\times 10^{\tiny{-7}}\text{ \frac{m.}{s.}}$ (a2) $|\bar{v}|=\frac{100}{1}\text{ \frac{ft.}{yr.}}\cdot \frac{0.3048\text{ m.}}{1\text{ ft.}}\cdot \frac{1\text{ yr.}}{31556925.9936\text{ s.}}\approx 9.659\times 10^{\tiny{-7}}\text{ \frac{m.}{s.}}$ (b) We can use $t=\frac{x}{|\bar{v}|}$, where time is t, average speed is $|\bar{v}|$ and displacement is x. We are given x = 3000 miles, and: $|\bar{v}|=10\text{ \frac{mm.}{yr.}}\cdot \frac{1\text{ m.}}{1000\text{ mm.}}\cdot \frac{0.3048\text{ ft.}}{1\text{ m.}}\cdot\frac{1\text{ mi.}}{5280\text{ ft.}}\approx 5.773\times 10^{\tiny{-7}}\text{ \frac{mi.}{yr.}}$ $t=\frac{3000\text{ mi.}}{5.773\times 10^{\tiny{-7}}\text{ \frac{mi.}{yr.}}}\approx 5.197\times 10^{\tiny{9}}\text{ yr.}$ 2. (a) $|\bar{v}|=\frac{20}{4000}=\frac{1}{200}\text{ \frac{ft.}{yr.}}\cdot \frac{0.3048\text{ m.}}{1\text{ ft.}}\cdot \frac{1\text{ yr.}}{31556925.9936\text{ s.}}\approx 4.83\times 10^{\tiny{-11}}\text{ \frac{m.}{s.}}$ (b) $|\bar{v}|=2\text{ \frac{ft.}{day}}\cdot\frac{0.3048\text{ m.}}{1\text{ ft.}}\cdot\frac{1\text{ day}}{86400\text{ s.}}\approx 7.06\times 10^{\tiny{-6}}\text{ \frac{m.}{s.}}$ 3. (a) The slower car takes $t_s=\frac{10\text{ mi.}}{55\text{ \frac{mi.}{hr.}}}=\frac{2}{11}\text{ hr.}$ to travel 10 miles, whereas the faster car takes $t_f=\frac{10\text{ mi.}}{70\text{ \frac{mi.}{hr.}}}=\frac{1}{7}\text{ hr.}$. To find how much sooner the faster car arrives than the slower car, we find the difference in their times. $\left(\frac{2}{11}-\frac{1}{7}\right)\text{ hr.}=\frac{3}{77}\text{ hr.}\approx 2\text{ min. }20.26\text{ s.}$ (b) 15 minutes is .25 hour. So we can use: $\frac{1}{4}\text{ hr.}=t_s-t_f=\frac{x\text{ mi.}}{55\text{ \frac{mi.}{hr.}}}-\frac{x\text{ mi.}}{70\text{ \frac{mi.}{hr.}}}\:\therefore\=\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: $\frac{\frac{3}{77}}{10}=\frac{\frac{1}{4}}{x}$ Cross multiply: $\frac{3x}{77}=\frac{5}{2}$ Multiply through by $\frac{77}{3}$: $x=\frac{385}{6}$

 Tags desert, dunes, island, move, physicis, problem, san, sand

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# sand dunes in a desert move over time as sand is swept up the windward side to settle in the lee side.

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post uzurpatorul Math Events 8 April 26th, 2017 04:15 PM drunkd Algebra 1 July 30th, 2012 12:27 PM r-soy Physics 6 November 4th, 2010 11:36 AM r-soy Physics 5 November 4th, 2010 10:47 AM r-soy Physics 2 October 29th, 2010 03:41 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top