b
(b) According to given problem \(\frac{1}{2}m{v^2} = a{s^2}\) \( \Rightarrow v = s\sqrt {\frac{{2a}}{m}} \)

So \({a_R} = \frac{{{v^2}}}{R} = \frac{{2a{s^2}}}{{mR}}\)…\((i)\)

Further more as \({a_t} = \frac{{dv}}{{dt}} = \frac{{dv}}{{ds}} \cdot \frac{{ds}}{{dt}} = v\frac{{dv}}{{ds}}\)…\((ii)\)   (By chain rule)

Which in light of equation \((i)\) i.e. \(v = s\sqrt {\frac{{2a}}{m}} \) yields

\({a_t} = \left[ {s\sqrt {\frac{{2a}}{m}} } \right]\,\left[ {\sqrt {\frac{{2a}}{m}} } \right] = \frac{{2as}}{m}\)…\((iii)\)

So that \(a = \sqrt {a_R^2 + a_t^2} = \sqrt {{{\left[ {\frac{{2a{s^2}}}{{mR}}} \right]}^2} + {{\left[ {\frac{{2as}}{m}} \right]}^2}} \)

Hence \(a = \frac{{2as}}{m}\sqrt {1 + {{[s/R]}^2}} \)

 \(F = ma = 2as\sqrt {1 + {{[s/R]}^2}} \)