A) Zero
B) \[\frac{1}{\sqrt{2}}\,\,m\text{/}{{s}^{\text{2}}}\] toward north-west
C) \[\frac{1}{\sqrt{2}}\,\,m\text{/}{{s}^{\text{2}}}\] toward north-east
D) \[\frac{1}{2}\,\,m\text{/}{{s}^{\text{2}}}\]toward north-west
Correct Answer: B
Solution :
\[\Delta \vec{\upsilon }={{\vec{\upsilon }}_{2}}-{{\vec{\upsilon }}_{1}}\]\[=\sqrt{\upsilon _{1}^{2}+\upsilon _{2}^{2}-2{{\upsilon }_{1}}{{\upsilon }_{2}}\,\,\cos {{90}^{o}}}\] \[=\sqrt{{{5}^{2}}+{{5}^{2}}}=5\sqrt{2}\] 
Average acceleration \[=\frac{\Delta \upsilon }{\Delta t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}\,\,\text{m/}{{\text{s}}^{\text{2}}}\] Directed toward north-west (As clear from the figure).
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