High School Physics · Grades 10–12

24Kinematics in Two Dimensions

Watch a car take a roundabout at a steady $30\,\mathrm{km}/\mathrm{h}$ : the needle never moves, yet every passenger feels pulled sideways — the motion is changing, and it is not the speed. One number per instant no longer suffices: motion in the plane needs arrows. This chapter equips a moving point with three — position, velocity, acceleration — chained by the derivative, and reads whole families of motion off their geometry.

24.1 The position vector

Motion only exists relative to a stated reference frame — a rigid body plus a clock (Chapter 5). New this year is precision: attach to the frame an origin $O$ and two perpendicular graduated axes $Ox$ , $Oy$ , and record where the point is at every instant, as a pair of functions.

Definition 24.1 (Position vector)

In a frame equipped with an origin $O$ and axes $Ox$ , $Oy$ , the position vector of a moving point $M$ at time $t$ is $\vect{OM}(t)$ , of coordinates $\bigl(x(t),\, y(t)\bigr)$ : two functions of time, one per axis. The curve traced by $M$ as $t$ runs is its trajectory.

24.2 The velocity vector

Definition 24.2 (Velocity vector)

Over a short interval the point moves by $\vect{OM}(t+\Delta t) - \vect{OM}(t)$ ; divide by $\Delta t$ and let the interval shrink: this is the derivative from your mathematics course, applied to each coordinate. The velocity vector of $M$ is the derivative of its position vector, $\vect v(t) = \bigl(x'(t),\, y'(t)\bigr)$ , in $\mathrm{m}/\mathrm{s}$ . Its norm $v = \sqrt{x'^2 + y'^2}$ is the speed — the one number the speedometer shows.

Example 24.3 (A drone in level flight)

A drone flies with $x(t) = 4.0\,t$ and $y(t) = 3.0\,t$ (meters, seconds): trajectory the line $y = \tfrac34 x$ , velocity $\vect v = (4.0, 3.0)$ at all times, speed $\sqrt{4.0^2 + 3.0^2} = 5.0\,\mathrm{m}/\mathrm{s}$ , constant.

Proposition 24.4 (Tangent to the trajectory)

At every instant, $\vect v(t)$ is tangent to the trajectory at the current position of $M$ , and points in the direction of travel.

Proof. The displacement is a chord of the trajectory; as $\Delta t$ shrinks, the chord’s direction becomes the tangent’s. ∎

24.3 The acceleration vector

Definition 24.5 (Acceleration vector)

The acceleration vector of $M$ is the derivative of its velocity vector, $\vect a(t) = \bigl(v_x'(t),\, v_y'(t)\bigr) = \bigl(x''(t),\, y''(t)\bigr)$ , in $\mathrm{m}/\mathrm{s}^{2}$ : how fast, and in which direction, the velocity vector is currently changing.

Remark 24.6 (Accelerating without speeding up)

$\vect a \neq \vect 0$ whenever the velocity vector changes — in length or in direction: a braking car accelerates (backward); a turning car at constant speed accelerates (sideways). This is the quantity forces control (Chapters 16 and 25).

Method 24.7 (Reading a chronophotograph)

A chronophotograph records the positions $M_0, M_1, M_2, \dots$ of a moving point at equal time intervals $\Delta t$ .

Dot spacing reads speed: equal spacings, constant speed; opening, speeding up; closing, slowing down.
The velocity at $M_i$ is the symmetric chord, $\vect v_i \approx \vect{M_{i-1}M_{i+1}}/(2\,\Delta t)$ , drawn at $M_i$ , tangent to the dotted path.
For the acceleration at $M_i$ : copy $\vect v_{i-1}$ and $\vect v_{i+1}$ to $M_i$ , draw $\Delta\vect v = \vect v_{i+1} - \vect v_{i-1}$ tip to tip; then $\vect a_i \approx \Delta\vect v/(2\,\Delta t)$ .

Chronophotograph of a thrown ball: v_2 and v_4, built from chords, are copied to M_3; their difference v hands over a — vertical, downward. — Chronophotograph of a thrown ball: $\vect v_2$ and $\vect v_4$ , built from chords, are copied to $M_3$ ; their difference $\Delta\vect v$ hands over $\vect a$ — vertical, downward.

24.4 Rectilinear motions

Along a straight line one axis suffices: $x(t)$ , $v(t) = x'(t)$ , $a(t) = v'(t)$ , all signed numbers.

Definition 24.8 (Uniform rectilinear motion)

A motion is uniform rectilinear when $\vect v$ is a constant vector: straight trajectory, constant speed, $\vect a = \vect 0$ . Then $x(t) = v\,t + x_0$ .

Proposition 24.9 (Uniformly accelerated rectilinear motion)

If a point moves along $Ox$ with constant acceleration $a$ , starting at $t = 0$ from position $x_0$ with velocity $v_0$ , then

v(t) = a\,t + v_0, \qquad x(t) = \tfrac12\,a\,t^2 + v_0\,t + x_0 .

Proof. $v$ is an antiderivative of the constant $a$ : $v(t) = a t + C$ , and $t = 0$ forces $C = v_0$ . In turn $x$ is an antiderivative of $a t + v_0$ : $x(t) = \tfrac12 a t^2 + v_0 t + C'$ , with $C' = x_0$ . ∎

Uniformly accelerated motion (a = 2.0\, m/ s2, v_0 = 1.0\, m/ s, x_0 = 0): each graph is the derivative of the one above — a parabola, its slope, the slope’s slope. — Uniformly accelerated motion ( $a = 2.0\,\mathrm{m}/\mathrm{s}^{2}$ , $v_0 = 1.0\,\mathrm{m}/\mathrm{s}$ , $x_0 = 0$ ): each graph is the derivative of the one above — a parabola, its slope, the slope’s slope.

Example 24.10 (Braking distance)

A car at speed $v_0$ brakes with constant deceleration $a$ (acceleration $-a$ ). It stops when $v(t) = -a t + v_0 = 0$ , at $t_s = v_0/a$ , having covered $d = -\tfrac12 a t_s^2 + v_0 t_s = v_0^2/(2a)$ . The square is the road-safety headline: doubling the speed quadruples the braking distance — with $a = 6.0\,\mathrm{m}/\mathrm{s}^{2}$ , $16\,\mathrm{m}$ from $50\,\mathrm{km}/\mathrm{h}$ , $64\,\mathrm{m}$ from $100\,\mathrm{km}/\mathrm{h}$ .

24.5 Uniform circular motion and the Frenet frame

Definition 24.11 (Uniform circular motion)

A motion is uniform circular when the trajectory is a circle of radius $R$ and the speed $v$ is constant. The period $T$ is the duration of one lap, and the frequency $f = 1/T$ (in $\mathrm{Hz}$ ) the number of laps per second: $v = 2\pi R/T = 2\pi R f$ .

Constant speed — yet the velocity turns: there is an acceleration. Which way, and how big?

Theorem 24.12 (Centripetal acceleration)

In uniform circular motion of radius $R$ and speed $v$ , the acceleration — called centripetal — points from the moving point toward the center of the circle, with constant norm

a = \frac{v^2}{R}.

Proof. Center the axes at the circle’s center and set $\omega = v/R$ , so the arc $v t$ subtends the angle $\omega t$ : the position is $x(t) = R\cos(\omega t)$ , $y(t) = R\sin(\omega t)$ . Differentiating, $\vect v = (-R\omega\sin(\omega t),\, R\omega\cos(\omega t))$ , of constant norm $R\omega = v$ ; differentiating again, $\vect a = -\omega^2\,\vect{OM}$ : opposite to the position vector — toward the center — of norm $\omega^2 R = v^2/R$ . ∎

Uniform circular motion: v tangent, of fixed length; a aimed at the center, of fixed length v2/R — forever turning the velocity without ever stretching it. — Uniform circular motion: $\vect v$ tangent, of fixed length; $\vect a$ aimed at the center, of fixed length $v^2/R$ — forever turning the velocity without ever stretching it.

Definition 24.13 (Frenet frame)

The Frenet frame at the current position of a point on a curved path is the pair of perpendicular unit vectors $(\vect{u_t}, \vect{u_n})$ : $\vect{u_t}$ tangent, along the motion; $\vect{u_n}$ normal, toward the center of the circle of radius $R$ that best fits the curve there.

Proposition 24.14 (Acceleration in the Frenet frame)

Along any plane path of local radius $R$ , traveled at speed $v(t)$ ,

\vect a = a_t\,\vect{u_t} + a_n\,\vect{u_n}, \qquad a_t = \frac{\dd v}{\dd t}, \qquad a_n = \frac{v^2}{R}:

the tangential acceleration $a_t$ changes the speed; the normal acceleration $a_n$ turns the velocity.

Proof. Admitted at this level. ∎

Remark 24.15 (Where the proof lives)

We proved the extremes: pure $a_t$ on a line (Proposition 24.9), pure $a_n$ on a circle (Theorem 24.12); the general decomposition is derived in the Year 1 volume.

Example 24.16 (Braking into a bend)

A car enters a bend of radius $R = 50\,\mathrm{m}$ at $v = 15\,\mathrm{m}/\mathrm{s}$ while braking at $\dd v/\dd t = -3.0\,\mathrm{m}/\mathrm{s}^{2}$ : $a_t = -3.0\,\mathrm{m}/\mathrm{s}^{2}$ , $a_n = 15^2/50 = 4.5\,\mathrm{m}/\mathrm{s}^{2}$ , so $a = \sqrt{3.0^2 + 4.5^2} \approx 5.4\,\mathrm{m}/\mathrm{s}^{2}$ , pointing backward and into the bend. The tires must supply both at once — why racing drivers brake before the turn.

24.6 Exercises

Exercise 24.1 ★

For $x(t) = 3.0\,t$ , $y(t) = 4.0\,t$ (meters, seconds), give $\vect v$ , the speed, the trajectory and the class of motion.

Solution

Solution of Exercise 24.1.

$\vect v = (3.0, 4.0)$ , constant; speed $\sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{m}/\mathrm{s}$ . Trajectory: the line $y = \tfrac43 x$ . Constant velocity vector: uniform rectilinear motion.

Exercise 24.2 ★

On a chronophotograph taken every $0.10\,\mathrm{s}$ , the dots lie on a line, evenly spaced $0.35\,\mathrm{m}$ apart. Identify the motion and compute the speed. What would opening spacings indicate?

Solution

Solution of Exercise 24.2.

Straight line, equal spacings: uniform rectilinear; $v = 0.35/0.10 = 3.5\,\mathrm{m}/\mathrm{s}$ . Opening spacings would mean the speed is increasing.

Exercise 24.3 ★

For $x(t) = 2.0\,t^2$ along $Ox$ (meters, seconds), derive $v(t)$ and $a(t)$ , name the class of motion, and compute the speed at $t = 3.0\,\mathrm{s}$ .

Solution

Solution of Exercise 24.3.

$v(t) = 4.0\,t$ , $a = 4.0\,\mathrm{m}/\mathrm{s}^{2}$ : constant acceleration, uniformly accelerated rectilinear. $v(3.0) = 12\,\mathrm{m}/\mathrm{s}$ .

Exercise 24.4 ★

A carousel horse rides a circle of radius $4.5\,\mathrm{m}$ , one lap every $8.0\,\mathrm{s}$ : compute the frequency, the speed and the acceleration (norm and direction).

Solution

Solution of Exercise 24.4.

$f = 1/8.0 = 0.125\,\mathrm{Hz}$ ; $v = 2\pi \times 4.5/8.0 \approx 3.5\,\mathrm{m}/\mathrm{s}$ ; $a = v^2/R = 3.53^2/4.5 \approx 2.8\,\mathrm{m}/\mathrm{s}^{2}$ , pointing toward the carousel’s axis.

Exercise 24.5 ★

A $v(t)$ graph climbs straight from $2.0\,\mathrm{m}/\mathrm{s}$ at $t = 0$ to $10.0\,\mathrm{m}/\mathrm{s}$ at $t = 4.0\,\mathrm{s}$ . Read off the acceleration, then the distance covered (area under the graph).

Solution

Solution of Exercise 24.5.

Slope: $a = (10.0 - 2.0)/4.0 = 2.0\,\mathrm{m}/\mathrm{s}^{2}$ . Area (trapezoid): $d = \tfrac12 (2.0 + 10.0) \times 4.0 = 24\,\mathrm{m}$ .

Exercise 24.6 ★★

A high-speed train takes a curve of radius $6.0\,\mathrm{km}$ at $320\,\mathrm{km}/\mathrm{h}$ . Compute $a_n$ and compare it to $g$ ; why must high-speed lines avoid tight curves, even perfectly banked ones?

Solution

Solution of Exercise 24.6.

$v = 88.9\,\mathrm{m}/\mathrm{s}$ : $a_n = 88.9^2/6000 \approx 1.3\,\mathrm{m}/\mathrm{s}^{2} \approx 0.13\,g$ . Since $a_n = v^2/R$ , high $v$ demands huge $R$ to keep the sideways acceleration (and the force on track and passengers) acceptable — banking changes who supplies it, not its size.

Exercise 24.7 ★★

A trolley on a straight rail has $x(t) = -1.2\,t^2 + 8.0\,t + 3.0$ (meters, seconds). Find $v(t)$ and $a$ , the instant and position of reversal, and describe the motion before and after.

Solution

Solution of Exercise 24.7.

$v(t) = -2.4\,t + 8.0$ ; $a = -2.4\,\mathrm{m}/\mathrm{s}^{2}$ , constant. Reversal at $v = 0$ : $t = 8.0/2.4 \approx 3.3\,\mathrm{s}$ , at $x \approx 16.3\,\mathrm{m}$ . Before: forward, slowing; after: backward, speeding up — uniformly accelerated throughout.

Exercise 24.8 ★★

A chronophotograph along a line, every $\Delta t = 0.20\,\mathrm{s}$ , gives $x = 0$ , $0.10$ , $0.28$ , $0.54$ , $0.88\,\mathrm{m}$ . Estimate the velocity at the three interior dots (Method 24.7), then the acceleration; conclude.

Solution

Solution of Exercise 24.8.

$v_1 = 0.28/0.40 = 0.70\,\mathrm{m}/\mathrm{s}$ ; $v_2 = (0.54 - 0.10)/0.40 = 1.10\,\mathrm{m}/\mathrm{s}$ ; $v_3 = (0.88 - 0.28)/0.40 = 1.50\,\mathrm{m}/\mathrm{s}$ . Velocity gains $0.40\,\mathrm{m}/\mathrm{s}$ every $0.20\,\mathrm{s}$ : $a = 2.0\,\mathrm{m}/\mathrm{s}^{2}$ , constant — uniformly accelerated rectilinear.

Exercise 24.9 ★★

The Moon orbits the Earth on a near circle of radius $3.84 \times 10^{8}\,\mathrm{m}$ in $27.3\,\mathrm{d}$ . Compute its speed and its centripetal acceleration; compare the latter to $g$ , knowing the Moon is about $60$ Earth radii away (Chapter 4).

Solution

Solution of Exercise 24.9.

$T = 2.36 \times 10^{6}\,\mathrm{s}$ : $v = 2\pi \times 3.84 \times 10^{8}/2.36 \times 10^{6} \approx 1.0 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ ; $a_n = v^2/R \approx 2.7 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ . Ratio: $g/a_n \approx 3600 = 60^2$ — gravity, diluted as the inverse square of distance (Chapter 4), exactly accounts for the Moon’s turning.

Exercise 24.10 ★★

A washing-machine drum of radius $25\,\mathrm{cm}$ spins at $1200$ revolutions per minute. Compute $f$ , $T$ , the rim speed and the rim acceleration in multiples of $g$ . Why does the laundry stay pinned while the water leaves?

Solution

Solution of Exercise 24.10.

$f = 1200/60 = 20\,\mathrm{Hz}$ , $T = 0.050\,\mathrm{s}$ ; $v = 2\pi R f = 2\pi \times 0.25 \times 20 \approx 31\,\mathrm{m}/\mathrm{s}$ ; $a_n = v^2/R \approx 3.9 \times 10^{3}\,\mathrm{m}/\mathrm{s}^{2} \approx 400\,g$ . The drum wall supplies the laundry’s $a_n$ ; through the holes nothing pushes the water inward, so it flies off along the tangent.

Exercise 24.11 ★★

A car at $90\,\mathrm{km}/\mathrm{h}$ brakes with constant $a = 7.5\,\mathrm{m}/\mathrm{s}^{2}$ : compute the stopping time and the braking distance, and sketch $v(t)$ .

Solution

Solution of Exercise 24.11.

$v_0 = 25\,\mathrm{m}/\mathrm{s}$ : $t_s = 25/7.5 \approx 3.3\,\mathrm{s}$ ; $d = v_0^2/(2a) = 625/15 \approx 42\,\mathrm{m}$ . Graph: a straight segment from $25\,\mathrm{m}/\mathrm{s}$ down to $0$ at $t_s$ .

Exercise 24.12 ★★★

For $x(t) = R\cos(\omega t)$ , $y(t) = R\sin(\omega t)$ : compute $\vect v$ ; check the speed is the constant $R\omega$ and $\vect v \perp \vect{OM}$ ; compute $\vect a$ , show it is $-\omega^2\,\vect{OM}$ , and recover $a = v^2/R$ .

Solution

Solution of Exercise 24.12.

$\vect v = (-R\omega\sin\omega t,\, R\omega\cos\omega t)$ , of constant norm $R\omega$ ; its dot product with $\vect{OM}$ vanishes, so it is perpendicular to the radius, hence tangent. Differentiating again, $\vect a = -\omega^2\vect{OM}$ : toward $O$ , of norm $\omega^2 R = (v/R)^2 R = v^2/R$ .

Exercise 24.13 ★★★

A motorcycle enters a bend of radius $80\,\mathrm{m}$ at $20\,\mathrm{m}/\mathrm{s}$ , braking at $2.5\,\mathrm{m}/\mathrm{s}^{2}$ . Compute $a_t$ , $a_n$ , the norm of $\vect a$ and its angle with the direction of motion. The tires grip up to $7.0\,\mathrm{m}/\mathrm{s}^{2}$ in all: is the rider within the limit?

Solution

Solution of Exercise 24.13.

$a_t = 2.5\,\mathrm{m}/\mathrm{s}^{2}$ (backward), $a_n = 20^2/80 = 5.0\,\mathrm{m}/\mathrm{s}^{2}$ ; $a = \sqrt{2.5^2 + 5.0^2} \approx 5.6\,\mathrm{m}/\mathrm{s}^{2}$ ; angle with the direction of motion: $\theta = 180^\circ - \arctan(5.0/2.5) \approx 117^\circ$ (i.e. $63^\circ$ from straight backward). $5.6 < 7.0$ : within grip, with little to spare.

Exercise 24.14 ★★★

A metro accelerates from rest at $1.2\,\mathrm{m}/\mathrm{s}^{2}$ up to $20\,\mathrm{m}/\mathrm{s}$ , cruises for $40\,\mathrm{s}$ , then brakes at $1.5\,\mathrm{m}/\mathrm{s}^{2}$ to a stop. Compute the duration and length of each phase, the total distance and the average speed. Sketch $v(t)$ and check the distance as an area.

Solution

Solution of Exercise 24.14.

Launch: $t_1 = 20/1.2 \approx 17\,\mathrm{s}$ , $d_1 = 20^2/(2 \times 1.2) \approx 167\,\mathrm{m}$ . Cruise: $40\,\mathrm{s}$ , $800\,\mathrm{m}$ . Braking: $t_3 = 20/1.5 \approx 13\,\mathrm{s}$ , $d_3 = 400/3.0 \approx 133\,\mathrm{m}$ . Total: $1.1 \times 10^{3}\,\mathrm{m}$ in $70\,\mathrm{s}$ ; average $1100/70 \approx 16\,\mathrm{m}/\mathrm{s}$ ( $57\,\mathrm{km}/\mathrm{h}$ ). $v(t)$ is a trapezoid whose area, $\tfrac12(70 + 40) \times 20 = 1100\,\mathrm{m}$ , checks the total.

Exercise 24.15 ★★★

A geostationary satellite circles at radius $4.22 \times 10^{7}\,\mathrm{m}$ from the Earth’s center with period $86\,164\,\mathrm{s}$ . Compute its speed and its centripetal acceleration. What supplies it, and why does the satellite hang over one fixed equatorial point (Chapter 27)?

Solution

Solution of Exercise 24.15.

$v = 2\pi \times 4.22 \times 10^{7}/86\,164 \approx 3.1 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ ; $a_n = v^2/R \approx 0.22\,\mathrm{m}/\mathrm{s}^{2}$ . Earth’s gravity at that distance supplies it; the period equals Earth’s rotation period, so satellite and ground turn together and it hangs over one equatorial point (Chapter 27).

24.7 Problem: The Accident Report

Problem 24.1

Weekend problem — the accident report: a dashcam film read frame by frame, a braking law rebuilt from skid marks, a roundabout that testifies about grip, and a verdict in kilometers per hour

A car strikes a van that pulled out of a side street; nobody is hurt, but the insurers disagree. The investigator has overhead camera footage (one frame every $0.50\,\mathrm{s}$ ), measured skid marks, and the car’s dashcam. The speed limit is $50\,\mathrm{km}/\mathrm{h}$ .

Part I — Reading the film. The frames before any reaction give, along the straight road:

$t$ ( $\mathrm{s}$ )	0.00	0.50	1.00	1.50	2.00
$x$ ( $\mathrm{m}$ )	0.0	9.0	18.0	27.0	36.0

Why must the report first name a frame and axes? Which are used here?
Compute the average velocity on each of the four intervals.
What class of motion is this? Justify from the table.
Give the car’s speed $v_0$ in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ .
What is the acceleration during these two seconds?

Part II — The skid marks. Tire tests on this dry asphalt give a constant braking deceleration $a = 6.0\,\mathrm{m}/\mathrm{s}^{2}$ ; the skid marks are $d_b = 27\,\mathrm{m}$ long. Take $t = 0$ , $x = 0$ at the start of braking, unknown speed $v_B$ .

By antiderivatives, write $v(t)$ and $x(t)$ during braking.
Express the stopping time $t_s$ ; show $d_b = v_B^2/(2a)$ .
Deduce $v_B$ from the skid marks. Consistent with Part I?
The reaction time is $1.0\,\mathrm{s}$ : compute the reaction distance and the total stopping distance.
Explain, from question 7, why doubling the speed quadruples the braking distance; compute $d_b$ at exactly $50\,\mathrm{km}/\mathrm{h}$ .

Part III — The roundabout testifies. Earlier, the dashcam shows the car rounding a roundabout lane of radius $R = 12.5\,\mathrm{m}$ at constant speed, a quarter turn in $2.5\,\mathrm{s}$ . The road can supply at most $a_n = \mu g$ , with $\mu = 0.70$ .

The speed is constant there — is the velocity? Is the car accelerating? Explain.
Compute the car’s speed on the roundabout.
Compute its centripetal acceleration (norm and direction).
Compute the maximum no-skid speed. Was the car within it?
Give the period and frequency of a full loop at that speed.

Part IV — The verdict. The van appeared $D = 41\,\mathrm{m}$ ahead of the car at the instant the driver started reacting.

By how much, in $\mathrm{km}/\mathrm{h}$ and in percent, did the speed before braking exceed the limit?
Compute the car’s speed at impact.
Redo the stopping computation at exactly $50\,\mathrm{km}/\mathrm{h}$ (same reaction, same braking): does the car stop, with what margin?
Give the general stopping-distance formula $d(v)$ ; why does speed enter twice — once linearly, once squared?
The verdict, one sentence with its numbers: what speed was the driver doing, and would the accident have happened at the limit?

Solution

Solution of Problem 24.1.

1. Motion is relative (Chapter 5): every position and velocity must name its frame. Here: the ground (road) frame, $x$ along the road from the first frame’s position.

2. Each interval: $9.0/0.50 = 18.0\,\mathrm{m}/\mathrm{s}$ — four equal values.

3. Straight road, equal displacements in equal times: uniform rectilinear motion.

4. $v_0 = 18.0\,\mathrm{m}/\mathrm{s} = 64.8\,\mathrm{km}/\mathrm{h}$ .

5. $\vect a = \vect 0$ .

6. $v(t) = -a\,t + v_B$ ; $x(t) = -\tfrac12 a t^2 + v_B\,t$ .

7. $v(t_s) = 0$ gives $t_s = v_B/a$ ; then $d_b = -\tfrac12 a (v_B/a)^2 + v_B^2/a = v_B^2/(2a)$ .

8. $v_B = \sqrt{2 a d_b} = \sqrt{2 \times 6.0 \times 27} = \sqrt{324} = 18.0\,\mathrm{m}/\mathrm{s}$ — exactly the filmed speed: the driver never slowed before braking.

9. Reaction: $18.0 \times 1.0 = 18\,\mathrm{m}$ ; total $18 + 27 = 45\,\mathrm{m}$ .

10. $d_b = v_B^2/(2a) \propto v_B^2$ : doubling $v_B$ multiplies $d_b$ by $4$ . At $50\,\mathrm{km}/\mathrm{h}$ ( $13.9\,\mathrm{m}/\mathrm{s}$ ): $d_b = 13.9^2/12.0 \approx 16\,\mathrm{m}$ .

11. The speed is constant but the velocity’s direction turns: $\vect v$ is not constant, so the car is accelerating.

12. Quarter turn: $2\pi R/4 = 19.6\,\mathrm{m}$ ; $v = 19.6/2.5 \approx 7.9\,\mathrm{m}/\mathrm{s}$ ( $28\,\mathrm{km}/\mathrm{h}$ ).

13. $a_n = 7.85^2/12.5 \approx 4.9\,\mathrm{m}/\mathrm{s}^{2}$ , toward the roundabout’s center.

14. $a_{\max} = \mu g = 0.70 \times 9.81 \approx 6.9\,\mathrm{m}/\mathrm{s}^{2}$ ; $v_{\max} = \sqrt{\mu g R} = \sqrt{6.87 \times 12.5} \approx 9.3\,\mathrm{m}/\mathrm{s}$ ( $33\,\mathrm{km}/\mathrm{h}$ ). $7.9 < 9.3$ : within grip — consistent with the clean, skid-free lane.

15. $T = 2\pi R/v = 78.5/7.85 \approx 10.0\,\mathrm{s}; f = 0.10\,\mathrm{Hz}$ .

16. $64.8 - 50 \approx 15\,\mathrm{km}/\mathrm{h}$ over — about $30\%$ above the limit.

17. Braking acts over $41 - 18 = 23\,\mathrm{m}$ : $v^2 = v_B^2 - 2ad = 324 - 2 \times 6.0 \times 23 = 48$ , so $v \approx 6.9\,\mathrm{m}/\mathrm{s} \approx 25\,\mathrm{km}/\mathrm{h}$ at impact.

18. At $13.9\,\mathrm{m}/\mathrm{s}$ : $13.9 \times 1.0 + 13.9^2/12.0 = 13.9 + 16.1 = 30.0\,\mathrm{m} < 41\,\mathrm{m}$ — the car stops $11\,\mathrm{m}$ short of the van.

19. $d(v) = v\,t_r + v^2/(2a)$ : once linearly (the reaction covers $v t_r$ before the brakes act), once squared (the kinetic part $v^2/(2a)$ ) — which is why moderate excesses cost disproportionately.

20. The dashcam and the skid marks agree: $65\,\mathrm{km}/\mathrm{h}$ in a $50\,\mathrm{km}/\mathrm{h}$ zone; at the legal limit the car would have stopped $11\,\mathrm{m}$ short of the van — the accident is the extra $15\,\mathrm{km}/\mathrm{h}$ .