area under the circle:
x2 + (y-a)2 = a2
[ acoswt=(...) , asinwt=(....) ... square n add.. ]
a point moves in the x-y plane according to the relation x=a sin wt and y=a(1-cos wt) The distance travelled by a point during time T is
Please give solution
area under the circle:
x2 + (y-a)2 = a2
[ acoswt=(...) , asinwt=(....) ... square n add.. ]
Vx = aw.coswt
Vy = aw.sinwt
=> V = a2w2
As velocity is constant... distance travelled in time T = V.T = a2w2.T...............
[1]
correct me if i'm wrong...
edited... [1]
sorry for such a silly mistake... [2] ...... actually i forgot to switch on my brain after waking up... [6] [4]
@neo.mind..
you have made a small mistake it will be underroot of that expression
and also take care that it is the speed that is constant not velocity.
hence distance traveled will be speed times. time.
[11]..bhaiya ,, mai kaha hu??..[11]
[12] [7] ...i mean..i 9 where am nw...but mai yaha kaha hu..[7]
hey neo.. sorry if it was not u.. but i think i replied to your post :)
May be it was mak's :)