A) \[3\sqrt{2}\]m/s northwards
B) \[\sqrt{2}\]m/s northwards
C) \[3\text{/}\sqrt{2}\]m/s northwards
D) None of the above
Correct Answer: C
Solution :
Here, Initial velocity \[{{v}_{1}}=15\,m\text{/}s\] (eastwards) Time taken \[t=10\,s\] and final velocity\[{{v}_{2}}=10\,m\text{/}s\](northwards) The change in velocity \[{{v}_{12}}=\sqrt{v_{1}^{2}+v_{2}^{2}-2{{v}_{1}}{{v}_{2}}\cos 90{}^\circ }\] \[=\sqrt{{{15}^{2}}+{{15}^{2}}}=15\sqrt{2}\] Hence, acceleration is \[a=\frac{{{v}_{12}}}{t}=\frac{15\sqrt{2}}{10}=\frac{3}{\sqrt{2}}\]northwards
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