Particle motion in polar coordinates

Given: $r = 2 + 0.5t^2$, $\theta = 0.4t^2$, at $t = 2$ s.

Derivatives at t = 2 s

Velocity

$$\begin{aligned} v_r &= \dot r = 2.00\ \text{m/s} \\ v_\theta &= r\dot\theta = 4 \times 1.6 = 6.40\ \text{m/s} \\ v &= \sqrt{2^2 + 6.4^2} = 6.71\ \text{m/s} \end{aligned}$$

Angle from the radial line: $\tan\phi = v_\theta / v_r = 6.4/2$, so $\phi = 72.6^\circ$.

Acceleration

$$\begin{aligned} a_r &= \ddot r - r\dot\theta^2 = 1 - 4(1.6)^2 = -9.24\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = 4(0.8) + 2(2)(1.6) = 9.60\ \text{m/s}^2 \\ a &= \sqrt{(-9.24)^2 + 9.6^2} = 13.32\ \text{m/s}^2 \end{aligned}$$

Since $a_r$ is negative and $a_\theta$ positive, the acceleration lies in the second quadrant of the $(e_r, e_\theta)$ axes, at $133.9^\circ$ measured from the $+e_r$ direction toward $+e_\theta$ (that is, $46.1^\circ$ from the $-e_r$ direction).

Answer: v = 6.71 m/s at 72.6° to the radial line; a = 13.32 m/s² at 133.9° to the radial line.