$\begin{array}{1 1} 2.39 \;m/s^2 \\ 1.39\;m/s^2 \\ 3.45\;m/s^2 \\1.98\;m/s^2\end{array} $

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Answer : $ 1.39\;m/s^2$

Here, $v_0 = 50 \;km/h =50 \times \large\frac{5}{18}$$m/s$

$\qquad= \large\frac{250}{18} $$m/s$ and

Since $a= \large\frac{v-v_0}{t}$

$\qquad= \large\frac {\Large\frac{300}{18}-\frac{250}{18}} {2}$

$\qquad= \large\frac{\Large\frac{50}{18}}{2}$

$\qquad= \large\frac{50}{36}$$=1.39 \;m/s^2$

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