Galilean Relativity and Tangential Accelerationby ek Tags: acceleration, galilean, relativity, tangential 

#1
Oct504, 10:49 PM

P: 186

Any help would be greatly appreciated.
1. Two swimmers, John and Emma, start together at the same point on the bank of a wide stream that flows with the speed c (c>v), relative to the water. John swims downstream a distance L and then upstream the same distance. Emma swims so that her motion relative to the earth is perpendicular to the banks of the stream. She swims the distance L and then back the same distance, so that both swimmers return to the starting point. Which swimmer returns first? 2. An automobile whose speed is increasing at a rate of .6 m/s^2 travels along a circular road of radius 20m. When the instantaneous speed of the automobile is 4 m/s, what is the tangential acceleration? Thanks. 



#2
Oct504, 10:59 PM

P: 1,445

what is v?
that is the same ether density determining experiment the swimmer going across would win 



#3
Oct604, 12:06 AM

P: 186

No v value given. Just asking you to solve in terms of time I suppose. Variables only.




#4
Oct604, 02:12 AM

P: 186

Galilean Relativity and Tangential Acceleration
Any help?
Please? 



#5
Oct604, 07:26 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

You were given help.
Her are more details: I assume that v is the speed of the water relative to the shore you should have said that. When John swims upstream his speed, relative to the bank, is c v because he is going against the water. How long will it take him to swim a distance L upstream? When John swims downstream, his speed, relative to the bank is c+ v because he is carried along by the water. How long will it take him to swim a distance L downstream? Add those to find the time required to swim both legs. Since Emma is swimming across the water, she needs to angle slightly up stream. Imagine drawing a line of length vt at an angle upstream, followed by a line of length ct straight down the stream back to the original horizontal. You get a right triangle with hypotenuse of length vt, one leg of length ct, and the other leg of length L, the distance she is swimming. By the Pythagorean theorem, (vt)^{2}= (ct)^{2}+ L^{2}. Solve that for t, the time required to swim the length L across the current. Because she comes back across the current, you can just double that to determine the time necessary to swim both legs. The second problem is pretty easy. If the speed were constant, the only acceleration would be toward the center of the circle there would be no "tangential acceleration". Since there is change in speed, the tangential acceleration is precisely that change in speed. 


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