The position vector of a particle which moves in the X-Y plane is given by \(\bar r = (3t^3 - 4t^2)\,i + (0.5t^4)\,j\) m. Calculate Velocity and acceleration at t = 4 Sec.

Velocity: \(\bar v = \dfrac{d\bar r}{dt} = (9t^2 - 8t)\,i + 2t^3\,j\)

At t = 4 s: \(\bar v = (144 - 32)\,i + 128\,j = 112\,i + 128\,j\) m/s

\(|v| = \sqrt{112^2 + 128^2} = \sqrt{28928} = 170.08\) m/s, inclined at \(\tan^{-1}(128/112) = 48.81^\circ\) to the X-axis.

Acceleration: \(\bar a = \dfrac{d\bar v}{dt} = (18t - 8)\,i + 6t^2\,j\)

At t = 4 s: \(\bar a = 64\,i + 96\,j\) m/s²; \(|a| = \sqrt{64^2 + 96^2} = \sqrt{13312} = 115.38\) m/s², at \(\tan^{-1}(96/64) = 56.31^\circ\) to the X-axis.

Answer: v = 112 i + 128 j m/s (170.08 m/s at 48.81°); a = 64 i + 96 j m/s² (115.38 m/s² at 56.31°).