11particle apni yatra shuru kahan se kar raha hai?????????????
try this problem
A particle moves in the plane xy with constant acceleration ω directed along the negative y axis. The equation of motion of the particle has the form y=ax-bx2, where a and b are positive constants. Find the velocity of the particle at the origin of coordinates.
this problem is very instresting...try it. i'l give solution afterwards.
-
UP 0 DOWN 0 0 6
6 Answers
106similarly,,, y'' is not acceleration nor is y' velocity as y'' = d2y/dx2 ≠d2t/dt2 and y'=dy/dx ≠dy/dt
1d velocity along x axis ll be √w/2b n for finding vy starting point of particle?
24arre haan..........main subah se maxima minima kar raha tha to dhyaan hi nahin raha.............thanx for correcting.....
106here accn in y axis = -w and along x-axis =0
y=ax-bx2
diff wrt time
Vy = aVx - 2bxVx ....... (i)
Diff again wrt time:
-w=-2bVx.Vx
==> Vx = √w/2b
Putting this value in (i) at (x,y)=(0,0)
Vy = a√w/2b
So, at (0,0) velocity of particle is
√w/2b i(cap) + a√w/2b j(cap)
Magnitude of velocity = √w(1+a2)/2b
hope its correct...
