\[(a)\;zero\;ms^{-2} \quad (b)\;20\;ms^{-2} \quad (c)\;\frac{10}{\sqrt 2} ms^{-2}\quad (d)\;10 \sqrt 2 ms^{-2}\]

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$acceleration=\large\frac{\overrightarrow v-\overrightarrow u}{t}$

$\qquad\qquad\quad=\large\frac{10 ms^{-1}\;East-10 m^{-1} due\; north}{1}$

$\qquad\qquad\quad=\large\frac{10 ms^{-1}\;East+10 m^{-1} due \;south}{1}$

$\qquad\qquad\quad=10 \sqrt 2 south\;east\;ms ^{-2}$

Hence d is the correct answer.

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