The position vector of a particle moving in the X–Y plane is \(\vec r=(2t^3-3t)\,\hat i+(t^2+4)\,\hat j\) m, where t is in seconds. At \(t=2\) s, find (i) the velocity and acceleration, (ii) the tangential and normal components of acceleration, and (…

Answer: \(\vec v=21\hat i+4\hat j\) (21.38 m/s), \(\vec a=24\hat i+2\hat j\) (24.08 m/s²); \(a_t=23.95\) m/s², \(a_n=2.53\) m/s²; \(\rho=180.9\) m.

(i) \(\vec v=\dfrac{d\vec r}{dt}=(6t^2-3)\hat i+2t\,\hat j\); \(\vec a=\dfrac{d\vec v}{dt}=12t\,\hat i+2\,\hat j\).

At \(t=2\) s: \(\vec v=21\hat i+4\hat j\), \(v=\sqrt{441+16}=\sqrt{457}=21.38\) m/s, at \(\tan^{-1}(4/21)=10.78^\circ\) to the x-axis.

\(\vec a=24\hat i+2\hat j\), \(a=\sqrt{576+4}=24.08\) m/s².

(ii) Tangential component (along \(\vec v\)):

\[a_t=\frac{\vec a\cdot\vec v}{v}=\frac{24(21)+2(4)}{21.38}=\frac{512}{21.38}=23.95\ \text{m/s}^2\]

Normal component:

\[a_n=\frac{|\vec v\times\vec a|}{v}=\frac{|21(2)-4(24)|}{21.38}=\frac{54}{21.38}=2.53\ \text{m/s}^2\]

Check: \(\sqrt{a_t^2+a_n^2}=\sqrt{573.6+6.4}=24.08\) m/s² \(=a\) ✓.

(iii) Radius of curvature: \(a_n=\dfrac{v^2}{\rho}\Rightarrow\rho=\dfrac{v^2}{a_n}=\dfrac{457}{2.526}=180.9\) m.