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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Jun 2007 19:37:57 IST
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A particle is moving in xy plane such that x=t + sint and y= cost. t is the time in sec.Find the length of the path taken by the partical from t=0 to t=2 sec.
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19 Jun 2007 20:35:54 IST
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x = ( t + sin t) i , y =( cos t ) j resultant = ( x^2 + y^2)^1/2 put values of t... to get the answer...(but first find t = 0 then t = 2pie)
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padhna likhna chad de mitra , nakal te rakhi aas , chak chaddar te so ja bhagta , Rabb karega tennu paas.....
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19 Jun 2007 20:43:43 IST
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the totle length covered is asked that is the distanance not displacement...
by ur method the displacement comes out.
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19 Jun 2007 20:47:21 IST
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The length of the path is asked, not the displacement.
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19 Jun 2007 20:49:37 IST
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yaar that u can never find in these type of qns. it is assumed that particle is moviong in a straight line...
also the topic is kinemetics....so it must be 1 - line motion i guess...
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padhna likhna chad de mitra , nakal te rakhi aas , chak chaddar te so ja bhagta , Rabb karega tennu paas.....
PLZ NEVER EVER RATE ME FOR MY ANSWERS , IF U WANT TO COMPLIMENT ME THEN JUST BEAT UR HEAD & SAY
"YEH KIS PAGAL NE MERA QN. SOLVE KER DIYA"
I AM SERIOUS!!!!
EVEN SERIOUS ^ INFINITY
PLZZZZZZ NEVER RATE ME....HOPE U WILL UNDERSTAND....
Shubham
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19 Jun 2007 20:57:57 IST
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x=t + sint
whn t = 0 , x = sin0 = 0
whn t = 2 pi, x= 2 +sin 2 pi =2
In this time intrval length =2
agn,
y= cost.
whn t = 0, y = 1
when t= 2 pi, y = cos 2 pi = 1
path in y dirn = 0
total path length = 2 units ?
p.s. there must be something else. it cant be that simple.//
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it is not important where u stand, but in which direction u are moving |
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19 Jun 2007 21:41:21 IST
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answer is 8...posting solution...
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Be Strong Be Different. Just Be
    
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19 Jun 2007 21:58:31 IST
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we have to find total length travelled as a function of t.... let the distance travelled in a small time dt be dl then the x and y coordinates change by dx and dy respectively as shown(both are non constant functions of t) therefore by pythagoras theorem the length dl =sqrt(dx2+dy2) but x=t+sint y=cost...differentiating....dx=(1+cost)dt dy=-sintdt...therefore dl=sqrt(dx2+dy2)=sqrt(((1+cost)dt)2+(sintdt)2)=sqrt(cos2t+sin2t+1+2cost)dt=sqrt(2+2cost)dt...therefore dl=sqrt(2+2cost)dt but the function changes sign ...therefore we have to find areas separately and add the moduli..(otherwise areas will get subtracted instead of added)... now integrating... = dl= 2sqrt(1+cost)dt= 2(cos(t/2))dt=4sin(t/2)+C...but C=0..now between t=0 to t= area =4 and between t= and t=2 the modulus of area is 4 therefore total area is 8.....now its right...
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Be Strong Be Different. Just Be
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20 Jun 2007 15:32:10 IST
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The length of the path is given by square root of (dx)^2+(dy)^2
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20 Jun 2007 20:45:36 IST
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whn t = 2 pi, x= 2 +sin 2 pi =2
error in this equation ramyani...
and of v follow ur sol v get 2pi so as im getting and many ppls r getting but ans is nt that ... i think ans is 2pi only... i have got wrongly printed ans i guess...
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20 Jun 2007 20:47:27 IST
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sorry rohit i wont vote u coz i too had got that ans...
well... thanks
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20 Jun 2007 21:52:06 IST
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ramyani plz be careful before u vote.. as u can see my previous answer was wrong...now i have changed it..and rini ur spelling of "KINEMATICS" is wrong..
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21 Jun 2007 16:14:34 IST
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Keep working....................Iam comming..
your's only,
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21 Jun 2007 16:17:34 IST
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yeah rohot....... u've got the correct answer...
thankyou... ( but still i wont vote u.... )
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21 Jun 2007 16:52:51 IST
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