The puck starts with velocity vector
[tex]\vec v_0=\left(2.35\dfrac{\rm m}{\rm s}\right)(\cos(-22^\circ)\,\vec\imath+\sin(-22^\circ)\,\vec\jmath)=(2.18\,\vec\imath-0.880\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]
Its velocity at time [tex]t[/tex] is
[tex]\vec v=\vec v_0+\vec at[/tex]
Over the 0.215 s interval, the velocity changes to
[tex]\vec v=\left(6.42\dfrac{\rm m}{\rm s}\right)(\cos50.0^\circ\,\vec\imath+\sin50.0^\circ\,\vec\jmath)=(4.13\,\vec\imath+4.92\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]
Then the acceleration must have been
[tex]\vec v=\vec v_0+(0.215\,\mathrm s)\vec a\implies\vec a=\dfrac{\vec v-\vec v_0}{0.215\,\rm s}=(9.06\,\vec\imath+27.0\,\vec\jmath)\dfrac{\rm m}{\mathrm s^2}[/tex]
which has a direction of about [tex]71.4^\circ[/tex].
