Given that the position vector of the moing ball at t th instant is given by

r(t)= t.e_{x- }sin(t).ey

Differentiating wrt t we get velocity vector

v(t)= r'(t)= 1.e_{x}-cos(t).e_{y}

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).e_{y}

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)