We can solve using the equation

\(v_{f}^{2}= v_{0}^{2}+2a\triangle x\)

where ?x is the distance over shich the accelerationoccurs. Solving this for a, we have \(\displaystyle{a}={\frac{{{v}^{{{2}}}_{\left\lbrace{f}\right\rbrace}-{v}^{{{2}}}_{\left\lbrace{0}\right\rbrace}}}{{{2}\triangle{x}}}}\)

We need to convert our velocity to units of m/s, so first wemultiply by 1000 m/km and divide by 3600 s/h which gives

vf= \(\displaystyle{\frac{{{\left({203}{k}\frac{{m}}{{h}}\right)}{\left({1000}\frac{{m}}{{k}}{m}\right)}}}{{{\left({3600}\frac{{s}}{{h}}\right)}}}}={56.4}\frac{{m}}{{s}}\)

Now, we just insert \(v_{f}= 56.4\) m/s, \(v_{0}= 0\),and \(\triangle x = 2000\)m into our equation, giving \(\displaystyle{a}={\frac{{{\left({56.4}\frac{{m}}{{s}}\right)}^{{{2}}}}}{{{2}{\left({2000}{m}\right)}}}}={0.79}\frac{{m}}{{s}^{{{2}}}}\)

so our answer is a) 0.79 m/s