a particle moves on the path y= sinx with constant speed u. what will be its acceleration at x= π/2
106y = sinx
=> V(y) = cosx*V(x)
V = √V2(x) + V2(y)
u = V(x)*√1+cos2x
=> V(x) = u/√1+cos2x
=> a(x) = u(1+cos2x)-1.5*sinx*cosx
at x = pi/2
a(x) = 0; and V(x) = u
=> a(y) = -sinx*V2(x) + cosx*a(x) (differentiating expression of V(y))
= -1*u2
So, |a| = u2
hence a = -u2 j
106alternatively, as speed is constant, tangential acceleration = 0.
Hence only acceleration is radial acc.
arad = v2/R = u2/R
R = radius of curvature at x = pi/2
Radius of curvature of a curve given by y = f(x) at a point (x,y) is given as
R = (1+(dydx)2)3/2d2ydx2
and now you can put x = pi/2, y=1 and evaluate the radius of curvature and hence calculate 'a'
