mechanics motion

y = 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

alternatively, as speed is constant, tangential acceleration = 0.

Hence only acceleration is radial acc.

a_rad = 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'

thanxx ashish

The second method is really nicer.!