Radar gun tracking car, polar components

Geometry (figure not shown): in the standard form of this problem the radar gun O lies on the circular road itself, and $\theta$ is measured from the diameter through O. The line from O to the car is then a chord of the circle of radius $R=250$ m:

$$r=2R\cos\theta=500\cos\theta$$

This is why the angle $\theta=40^\circ$ is given.

Derivatives of $r$

$$\dot r=-500\sin\theta\,\dot\theta,\qquad \ddot r=-500\left(\sin\theta\,\ddot\theta+\cos\theta\,\dot\theta^{2}\right)$$

At $\theta=40^\circ$ ($\cos40^\circ=0.7660$, $\sin40^\circ=0.6428$):

$$\begin{aligned} r&=500(0.7660)=383.02\ \text{m}\\ \dot r&=-500(0.6428)(0.15)=-48.21\ \text{m/s}\\ \ddot r&=-500\left[0.6428(0.025)+0.7660(0.15)^{2}\right]=-16.65\ \text{m/s}^2 \end{aligned}$$

Velocity

$$v_r=\dot r=-48.21\ \text{m/s},\qquad v_\theta=r\dot\theta=383.02(0.15)=57.45\ \text{m/s}$$ $$v=\sqrt{48.21^{2}+57.45^{2}}=75.0\ \text{m/s}$$

Acceleration

$$\begin{aligned} a_r&=\ddot r-r\dot\theta^{2}=-16.65-383.02(0.15)^{2}=-25.27\ \text{m/s}^2\\ a_\theta&=r\ddot\theta+2\dot r\dot\theta=383.02(0.025)+2(-48.21)(0.15)=-4.89\ \text{m/s}^2\\ a&=\sqrt{25.27^{2}+4.89^{2}}=25.74\ \text{m/s}^2 \end{aligned}$$

Check: the radius from the road's centre to the car turns through $2\theta$, so $v=R(2\dot\theta)=250(0.30)=75$ m/s, $a_t=R(2\ddot\theta)=12.5$ m/s$^2$, $a_n=v^2/R=22.5$ m/s$^2$ and $a=\sqrt{12.5^2+22.5^2}=25.74$ m/s$^2$.

(If the figure instead puts the gun at the centre of the road, $r=250$ m is constant and $v=r\dot\theta=37.5$ m/s, $a=\sqrt{(r\dot\theta^2)^2+(r\ddot\theta)^2}=8.41$ m/s$^2$.)

Answer: $v=75.0$ m/s; $a\approx25.7$ m/s$^2$.