---
title: "Kinematics in Two Dimensions"
book: "High School Physics"
subject: physics
language: en
chapter: 24
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions
---

# Chapter 24 — Kinematics 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](#def-g12-kinematics-2d-velocity). 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](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-frame) — a rigid body plus a clock ([Chapter 5](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#ch-g10-relative-motion)). 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](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-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](#def-g12-kinematics-2d-position), $\vect v(t) = \bigl(x'(t),\, y'(t)\bigr)$, 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](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) the line $y = \tfrac34 x$, velocity $\vect v = (4.0, 3.0)$ at all times, [speed](#def-g12-kinematics-2d-velocity) $\sqrt{4.0^2 + 3.0^2} = 5.0\,\mathrm{m}/\mathrm{s}$, constant.

**Proposition 24.4 (Tangent to the trajectory).**

At every instant, $\vect v(t)$ is tangent to the [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) at the current position of $M$, and points in the direction of travel.

**Proof.** The displacement is a chord of the [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-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](#def-g12-kinematics-2d-velocity), $\vect a(t) = \bigl(v_x'(t),\, v_y'(t)\bigr) =
\bigl(x''(t),\, y''(t)\bigr)$, in $\mathrm{m}/\mathrm{s}^{2}$: how fast, and in which direction, the [velocity vector](#def-g12-kinematics-2d-velocity) 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](#def-g12-kinematics-2d-velocity) accelerates (sideways). This is the quantity [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) control (Chapters [16](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#ch-g11-forces-and-motion) and [25](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws)).

**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 $\Delta t$.

1. Dot spacing reads [speed](#def-g12-kinematics-2d-velocity) : equal spacings, constant [speed](#def-g12-kinematics-2d-velocity) ; opening, speeding up; closing, slowing down.
2. 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.
3. 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-kinematics-2d/fig-294866202bd1.svg)

*[Chronophotograph](#met-g12-kinematics-2d-chrono) 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](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory), constant [speed](#def-g12-kinematics-2d-velocity), $\vect a = \vect 0$. 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $C = v_0$. 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-kinematics-2d/fig-ad4c6ed03948.svg)

*[Uniformly accelerated motion](#prop-g12-kinematics-2d-uarm) ($a = 2.0\,\mathrm{m}/\mathrm{s}^{2}$, $v_0 = 1.0\,\mathrm{m}/\mathrm{s}$, $x_0 = 0$): 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](#def-g12-kinematics-2d-velocity) $v_0$ 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](#def-g12-kinematics-2d-velocity) *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](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) is a circle of radius $R$ and the [speed](#def-g12-kinematics-2d-velocity) $v$ 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](#def-g12-kinematics-2d-velocity) — yet the velocity turns: there *is* an acceleration. Which way, and how big?

**Theorem 24.12 (Centripetal acceleration).**

In [uniform circular motion](#def-g12-kinematics-2d-ucm) of radius $R$ and [speed](#def-g12-kinematics-2d-velocity) $v$, 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](#def-g12-kinematics-2d-position) — 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-kinematics-2d/fig-213e819fb580.svg)

*[Uniform circular motion](#def-g12-kinematics-2d-ucm): $\vect v$ 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](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-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](#def-g12-kinematics-2d-velocity) $v(t)$,

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

the *tangential acceleration* $a_t$ changes the [speed](#def-g12-kinematics-2d-velocity); 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](#prop-g12-kinematics-2d-uarm)), pure $a_n$ on a circle ([Theorem 24.12](#thm-g12-kinematics-2d-centripetal)); 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](#def-g12-kinematics-2d-velocity), the [trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory) and the class of motion.

**Solution of Exercise 24.1.**

$\vect v = (3.0, 4.0)$, constant; [speed](#def-g12-kinematics-2d-velocity) $\sqrt{3.0^2 + 4.0^2} =
5.0\,\mathrm{m}/\mathrm{s}$. [Trajectory](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-trajectory): the line $y = \tfrac43 x$. Constant [velocity vector](#def-g12-kinematics-2d-velocity): [uniform rectilinear motion](#def-g12-kinematics-2d-uniform).

**Exercise 24.2 ★.**

On a [chronophotograph](#met-g12-kinematics-2d-chrono) 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](#def-g12-kinematics-2d-velocity). What would opening spacings indicate?

**Solution of Exercise 24.2.**

Straight line, equal spacings: [uniform rectilinear](#def-g12-kinematics-2d-uniform); $v = 0.35/0.10 = 3.5\,\mathrm{m}/\mathrm{s}$. Opening spacings would mean the [speed](#def-g12-kinematics-2d-velocity) 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](#def-g12-kinematics-2d-velocity) at $t = 3.0\,\mathrm{s}$.

**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](#def-g12-kinematics-2d-ucm), the [speed](#def-g12-kinematics-2d-velocity) and the acceleration (norm and direction).

**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 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](#def-g12-kinematics-2d-velocity) 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](#def-g12-kinematics-2d-velocity) lines avoid tight curves, even perfectly banked ones?

**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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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 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](#met-g12-kinematics-2d-chrono) 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](#met-g12-kinematics-2d-chrono)), then the acceleration; conclude.

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) the Earth on a near circle of radius $3.84 \times 10^{8}\,\mathrm{m}$ in $27.3\,\mathrm{d}$. Compute its [speed](#def-g12-kinematics-2d-velocity) and its [centripetal acceleration](#thm-g12-kinematics-2d-centripetal); compare the latter to $g$, knowing the Moon is about $60$ Earth radii away ([Chapter 4](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#ch-g10-universal-gravitation)).

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#ch-g10-universal-gravitation)), 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](#def-g12-kinematics-2d-velocity) and the rim acceleration in multiples of $g$. Why does the laundry stay pinned while the water leaves?

**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 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](#def-g12-kinematics-2d-velocity) 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 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 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](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#def-g10-relative-motion-speed). Sketch $v(t)$ and check the distance as an area.

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) circles at radius $4.22 \times 10^{7}\,\mathrm{m}$ from the Earth’s center with [period](#def-g12-kinematics-2d-ucm) $86\,164\,\mathrm{s}$. Compute its [speed](#def-g12-kinematics-2d-velocity) and its [centripetal acceleration](#thm-g12-kinematics-2d-centripetal). What supplies it, and why does the [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) hang over one fixed equatorial point ([Chapter 27](https://one-course.com/books/physics/2/en/chapter/27-satellites-and-planetary-motion#ch-g12-satellites-kepler))?

**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](#def-g12-kinematics-2d-ucm) equals Earth’s rotation [period](#def-g12-kinematics-2d-ucm), so [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) and ground turn together and it hangs over one equatorial point ([Chapter 27](https://one-course.com/books/physics/2/en/chapter/27-satellites-and-planetary-motion#ch-g12-satellites-kepler)).

## 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](#def-g12-kinematics-2d-velocity) 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 |

1. Why must the report first name a frame and axes? Which are used here?
2. Compute the average velocity on each of the four intervals.
3. What class of motion is this? Justify from the table.
4. Give the car’s [speed](#def-g12-kinematics-2d-velocity) $v_0$ in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ .
5. 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](#def-g12-kinematics-2d-velocity) $v_B$.

6. By antiderivatives, write $v(t)$ and $x(t)$ during braking.
7. Express the stopping time $t_s$ ; show $d_b = v_B^2/(2a)$ .
8. Deduce $v_B$ from the skid marks. Consistent with Part I?
9. The reaction time is $1.0\,\mathrm{s}$ : compute the reaction distance and the total stopping distance.
10. Explain, from question 7, why doubling the [speed](#def-g12-kinematics-2d-velocity) 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](#def-g12-kinematics-2d-velocity), a quarter turn in $2.5\,\mathrm{s}$. The road can supply at most $a_n = \mu g$, with $\mu = 0.70$.

11. The [speed](#def-g12-kinematics-2d-velocity) is constant there — is the velocity? Is the car accelerating? Explain.
12. Compute the car’s [speed](#def-g12-kinematics-2d-velocity) on the roundabout.
13. Compute its [centripetal acceleration](#thm-g12-kinematics-2d-centripetal) (norm and direction).
14. Compute the maximum no-skid [speed](#def-g12-kinematics-2d-velocity) . Was the car within it?
15. Give the [period](#def-g12-kinematics-2d-ucm) and [frequency](#def-g12-kinematics-2d-ucm) of a full loop at that [speed](#def-g12-kinematics-2d-velocity) .

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

16. By how much, in $\mathrm{km}/\mathrm{h}$ and in percent, did the [speed](#def-g12-kinematics-2d-velocity) before braking exceed the limit?
17. Compute the car’s [speed](#def-g12-kinematics-2d-velocity) at impact.
18. Redo the stopping computation at exactly $50\,\mathrm{km}/\mathrm{h}$ (same reaction, same braking): does the car stop, with what margin?
19. Give the general stopping-distance formula $d(v)$ ; why does [speed](#def-g12-kinematics-2d-velocity) enter twice — once linearly, once squared?
20. The verdict, one sentence with its numbers: what [speed](#def-g12-kinematics-2d-velocity) was the driver doing, and would the accident have happened at the limit?

**Solution of Problem 24.1.**

**1.** Motion is relative ([Chapter 5](https://one-course.com/books/physics/2/en/chapter/5-relative-motion#ch-g10-relative-motion)): 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](#def-g12-kinematics-2d-uniform).

**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](#def-g12-kinematics-2d-velocity): 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](#def-g12-kinematics-2d-velocity) 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 $15\,\mathrm{km}/\mathrm{h}$.