Given that the position vector of the moing ball at t th instant is given by
r(t)= t.ex- sin(t).ey
Differentiating wrt t we get velocity vector
v(t)= r'(t)= 1.ex-cos(t).ey
Hence the angle of velocity vector with the positive direction of x axis at t=1 is in 4th quadrant and the angle of inclination is
tan^-1cos(1)
Now the accleration at t th instant is
a(t)=v'(t)=sin(t).ey
Hence acceleration vector is directed towards +ve direction of y axis .
So angle beween accleration and velocity vector at t=1 is given by
π/2+tan^-1cos(1)