The motion of a particle is defined by the position vector, \(\bar r = 5t\,i + 3t^2\,j\) where ‘r’ is in meters and t is in seconds. At the instant when t = 2 sec, find i) Tangential and normal components of accelerations ii) Radius of curvature.

\(\bar v = 5\,i + 6t\,j\); at t = 2 s: \(\bar v = 5\,i + 12\,j\), \(v = \sqrt{25 + 144} = 13\) m/s

\(\bar a = 6\,j\) m/s² (constant), \(a = 6\) m/s²

i) Tangential component: \(a_t = \dfrac{\bar a \cdot \bar v}{v} = \dfrac{0 \times 5 + 6 \times 12}{13} = 5.538\) m/s²

Normal component: \(a_n = \dfrac{|\bar v \times \bar a|}{v} = \dfrac{|5 \times 6 - 12 \times 0|}{13} = 2.308\) m/s² (check: \(\sqrt{5.538^2 + 2.308^2} = 6\) ✔)

ii) Radius of curvature: \(\rho = \dfrac{v^2}{a_n} = \dfrac{169}{2.308} = 73.23\) m

Answer: \(a_t = 5.54\) m/s², \(a_n = 2.31\) m/s², \(\rho = 73.23\) m.