$\begin{gathered}
  \,\,\,\,\,Here,a = 15\,m{s^{ – 2}} \hfill \\
  \,\,\,\,\,R = 2.5\,m \hfill \\
  From\,figure, \hfill \\
  {a_c} = a\,\cos \,{30^ \circ } = 15 \times \frac{{\sqrt 3 }}{2}\,m{s^{ – 2}} \hfill \\
  As\,we\,know,\,{a_c} = \frac{{{v^2}}}{R} \Rightarrow v = \sqrt {{a_c}R}  \hfill \\
  \therefore \,\,\,\,\,v = \sqrt {15 \times \frac{{\sqrt 3 }}{2} \times 2.5}  = 5.69 = 5.7m\,{s^{ – 1}} \hfill \\ 
\end{gathered} $