---
title: "Radar gun tracking car, polar components"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:radar-gun-car-polar"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/radar-gun-car-polar
site: BCT Engineering
---

# Radar gun tracking car, polar components

Applied Mechanics (Dynamics) · Chapter 1: Curvilinear Motion of Particles

- Asked in: 2078 Bhadra, 4 marks
- Video: [Polar coordinates of car on circular track](https://www.youtube.com/watch?v=0lSXVZq0_Rw) — STEM Course Prep
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/radar-gun-car-polar
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

A radar gun at 'O' rotates with the angular velocity of $\dot\theta = 0.15$ rad/sec and angular acceleration of $\ddot\theta = 0.025$ rad/sec$^2$ at the instant $\theta = 40^\circ$, as it follows the motion of the car travelling along the circular road having radius of $r = 250$ m. Determine the magnitude of velocity and acceleration of the car at this instant. [Figure: gun at O, car on a circular road of radius 250 m at angle $\theta$.]

## Short answer

The distance from gun to car is r = 500 cosθ. Differentiating gives radial and transverse components; combining yields v = 75.0 m/s and a = 25.7 m/s².

## Answer

**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$^2$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is r_dot = 0 and r_double_dot = 0 for this car?
   A. The gun is at the centre of the circular road so r is constant
   B. The car moves at constant speed
   C. theta is constant
   D. The angle 40 degrees makes r constant
   Answer: A. With the gun at the centre, the radius to the car equals the road radius, 250 m, always. (Key point: Assumption)

2. What is the car's speed?
   A. r theta_dot = 250(0.15) = 37.5 m/s
   B. r theta_double_dot = 6.25 m/s
   C. theta_dot / r = 0.0006 m/s
   D. r theta_dot^2 = 5.625 m/s
   Answer: A. With r_dot = 0, v = v_theta = r theta_dot = 37.5 m/s. (Key point: Speed)

3. Which pair gives a_r and a_theta?
   A. a_r = -5.625, a_theta = 6.25 m/s^2
   B. a_r = 5.625, a_theta = 37.5 m/s^2
   C. a_r = 0, a_theta = 6.25 m/s^2
   D. a_r = -6.25, a_theta = 5.625 m/s^2
   Answer: A. a_r = -r theta_dot^2 = -250(0.0225) and a_theta = r theta_double_dot = 250(0.025). (Key point: Acceleration components)

4. Is the angle 40 degrees needed for the magnitudes?
   A. Yes, it enters v and a directly
   B. No, polar components depend only on r, theta_dot and theta_double_dot
   C. Yes, to compute r
   D. Yes, to find theta_dot
   Answer: B. Polar component formulas contain no theta itself, so the answer is 37.5 m/s and 8.41 m/s^2. (Key point: Role of the angle)

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