(b) $\Delta \vec \upsilon = {\vec \upsilon _2} – {\vec \upsilon _1}$$ = \sqrt {\upsilon _1^2 + \upsilon _2^2 – 2{\upsilon _1}{\upsilon _2}\,\,\cos {{90}^o}} $$ = \sqrt {{5^2} + {5^2}} = 5\sqrt 2 $

Average acceleration

$ = \frac{{\Delta \upsilon }}{{\Delta t}} = \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}\,\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

Directed toward north-west   (As clear from the figure).