Chapter 2 · Motion in a Straight Line

2.1 Position, Displacement and Instantaneous Velocity

Where a body is, how far it has really moved, and how fast it is moving at one instant: position, displacement and instantaneous velocity — the slope of the position–time graph, $v=\dfrac{dx}{dt}$.

A taxi takes you $24\ \text{km}$ from home to the airport in $40$ minutes, an average of $36\ \text{km/h}$. Yet the speedometer read $0$ at every traffic signal and touched $80\ \text{km/h}$ on the flyover. A speed camera on the flyover does not care about your average: it fines you for how fast you were going at that instant. And if you ended up back home because you forgot your passport, your average velocity for the whole trip would be zero, even though the meter charged you for every kilometre.

These everyday puzzles show that "how far" and "how fast" each have two meanings, and physics needs to keep them apart. In this section we describe straight-line motion by a position that can be positive or negative, separate path length from displacement, and then shrink the time interval until the average velocity becomes the instantaneous velocity — the number a speedometer shows. That last step is your first use of calculus in physics: velocity is the rate of change of position, $v=\dfrac{dx}{dt}$.

Position, frame of reference and displacement

To say where something is, you need a starting point, a direction and a clock. A motion in a straight line needs only one axis: we place it along the line of motion, call it the $x$-axis, and measure positions from a chosen origin $O$.

📘 Definition 2.1 — Frame of reference and position

A frame of reference is a coordinate system (an origin and axes) together with a clock, fixed to some chosen body — the ground, a train, the Earth's centre. For motion along a straight line we use one axis, the $x$-axis, along the line.

The position of a point object at time $t$ is its coordinate $x(t)$ on this axis: a distance from the origin with a sign, positive on the side chosen as the positive direction and negative on the other side. The object is at rest in the frame if $x$ does not change with time, and in motion if it does.

Rest and motion are relative: a passenger dozing in a train is at rest in the train's frame and moving at $90\ \text{km/h}$ in the ground's frame. Unless we say otherwise, the frame is fixed to the ground. We also treat the moving body as a point object — its size is small compared with the distance it moves — so a single number $x$ fixes where it is.

📘 Definition 2.2 — Displacement and path length

If an object moves from position $x_1$ at time $t_1$ to position $x_2$ at time $t_2$, its displacement is the change of position $$\Delta x=x_2-x_1 .$$ Displacement has a magnitude and a direction, shown by its sign: $\Delta x\gt0$ means a net shift in the $+x$ direction, $\Delta x\lt0$ in the $-x$ direction.

The path length (distance travelled) is the total length of the path actually covered. It is never negative, and $\text{path length}\ge|\Delta x|$, with equality only when the object never reverses direction.

Displacement depends only on where the motion starts and ends; path length depends on everything in between. Moving the origin changes every position but not any displacement, because both $x_1$ and $x_2$ shift by the same amount.

✏️ Example 1 — A jog along the beach

On a straight beach path, take the origin at a lifeguard tower and the $+x$ direction towards the pier. A jogger starts at $x=+50\ \text{m}$, runs to the pier at $x=+450\ \text{m}$, then turns and jogs back to an ice-cream stall at $x=+200\ \text{m}$. Find the displacement and the path length (a) for the outward run, (b) for the return leg, (c) for the whole jog. (d) What would they be if she jogged all the way back to her start?

💡 Show solution▶

(a) $\Delta x=450-50=+400\ \text{m}$ (towards the pier); path length $400\ \text{m}$.

(b) $\Delta x=200-450=-250\ \text{m}$ (towards the tower); path length $250\ \text{m}$ — always positive.

(d) Back at the start, $\Delta x=0$, but the path length is $400+400=800\ \text{m}$.

Check In (c) the displacements of the two legs add with their signs: $(+400)+(-250)=+150\ \text{m}$ ✓; the path lengths add without signs.

Average velocity and average speed

Dividing by the time taken turns each kind of "how far" into a kind of "how fast".

📘 Definition 2.3 — Average velocity and average speed

Over a time interval $\Delta t=t_2-t_1$, the average velocity and the average speed are

$$\bar v=\frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1},$$ $$\text{average speed}=\frac{\text{path length}}{\Delta t}.$$

Both have the SI unit $\text{m/s}$ and dimensions $[\text{L}\,\text{T}^{-1}]$. Average velocity carries a sign (a direction); average speed never does, and $|\bar v|\le$ average speed. On an x–t graph, $\bar v$ is the slope of the chord joining the points $(t_1,x_1)$ and $(t_2,x_2)$.

✏️ Example 2 — Averages for the jog

In Example 1 the outward run takes $100\ \text{s}$ and the return leg $75\ \text{s}$. Find the average velocity and average speed for each leg and for the whole jog.

💡 Show solution▶

Out $\bar v=\dfrac{+400\ \text{m}}{100\ \text{s}}=+4.0\ \text{m/s}$; average speed $4.0\ \text{m/s}$.

Back $\bar v=\dfrac{-250\ \text{m}}{75\ \text{s}}=-3.3\ \text{m/s}$; average speed $3.3\ \text{m/s}$.

Whole jog $\bar v=\dfrac{+150\ \text{m}}{175\ \text{s}}=+0.86\ \text{m/s}$ (towards the pier); average speed $=\dfrac{650\ \text{m}}{175\ \text{s}}=3.7\ \text{m/s}$.

Check $|\bar v|=0.86\ \text{m/s}\le3.7\ \text{m/s}$ ✓. Note that the whole-trip average speed is not the mean of $4.0$ and $3.3$: the legs take different times.

Instantaneous velocity: the slope of the x–t graph

An average hides what happens inside the interval. To find how fast the jogger is moving at a particular instant $t$, take the average over a shorter and shorter interval $\Delta t$ that starts at $t$. For smooth motion the averages settle down to a definite value. Geometrically, the chord joining $(t,x)$ to $(t+\Delta t,\,x+\Delta x)$ swings round until it lies along the tangent to the x–t curve at $t$.

📘 Definition 2.4 — Instantaneous velocity and speed

The instantaneous velocity at time $t$ is the limit of the average velocity as the interval shrinks to zero:

$$v=\lim_{\Delta t\to0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}.$$

$\dfrac{dx}{dt}$ is read "the derivative of $x$ with respect to $t$", or simply "the rate of change of position". It equals the slope of the tangent to the x–t graph at that instant. Its sign gives the direction of motion. The instantaneous speed is its magnitude, $|v|$.

From now on, "velocity" and "speed" without the word "average" always mean the instantaneous values. For motion in one direction at a steady rate — uniform motion — the x–t graph is a straight line, every chord has the same slope, and the instantaneous and average velocities are equal.

✏️ Example 3 — Shrinking the interval

A sprinter leaving the blocks has position $x=4.0\,t^2$ (in metres, $t$ in seconds) for the first two seconds. Find the average velocity over the intervals from $t=2.0\ \text{s}$ to $t=2.0\ \text{s}+\Delta t$ for $\Delta t=1$, $0.1$, $0.01$ and $0.001\ \text{s}$ (assume the formula holds a little beyond $2.0\ \text{s}$), and hence the velocity at $t=2.0\ \text{s}$.

💡 Show solution▶

General $\Delta x=4.0(2+\Delta t)^2-4.0(2)^2=16\,\Delta t+4.0\,(\Delta t)^2$, so $\dfrac{\Delta x}{\Delta t}=16+4.0\,\Delta t$.

Table $\Delta t=1$: $20\ \text{m/s}$; $\ 0.1$: $16.4\ \text{m/s}$; $\ 0.01$: $16.04\ \text{m/s}$; $\ 0.001$: $16.004\ \text{m/s}$.

Limit As $\Delta t\to0$ the term $4.0\,\Delta t$ vanishes: $v(2.0\ \text{s})=16\ \text{m/s}$.

Check Interactive 1 draws this chord shrinking onto the tangent; its slope readout settles on $16$ for $t=2$.

Repeating the algebra of Example 3 at a general time $t$ gives $\dfrac{\Delta x}{\Delta t}=8.0\,t+4.0\,\Delta t\to8.0\,t$, so $v=8.0\,t$. You do not have to redo the limit every time: a few rules, which you will prove in Class 11 Mathematics, cover all the motions in this chapter.

📐 Result 2.1 — Rates of change you will use

For constants $c$, $A$, $B$ and a whole number $n\ge1$ (you will prove these in maths):

$$\frac{d}{dt}(c)=0,\qquad \frac{d}{dt}\left(t^{n}\right)=n\,t^{\,n-1},$$ $$\frac{d}{dt}\big(A\,f+B\,g\big)=A\frac{df}{dt}+B\frac{dg}{dt}.$$

So for $x=x_0+bt+ct^2+dt^3$, the velocity is $v=\dfrac{dx}{dt}=b+2ct+3dt^2$. The initial position $x_0$ drops out: moving the origin never changes a velocity.

✏️ Example 4 — When does the particle turn back?

A particle moves along the $x$-axis with $x=t^3-6t^2+9t+2$ ($x$ in metres, $t$ in seconds). (a) Find its velocity as a function of time. (b) When is it momentarily at rest, and where? (c) Find its displacement and path length between $t=0$ and $t=4\ \text{s}$, and its average velocity and average speed.

💡 Show solution▶

(a) By Result 2.1, $v=\dfrac{dx}{dt}=3t^2-12t+9=3(t-1)(t-3)\ \text{m/s}$.

(b) $v=0$ at $t=1\ \text{s}$ (where $x=1-6+9+2=6\ \text{m}$) and at $t=3\ \text{s}$ (where $x=27-54+27+2=2\ \text{m}$). Between these times $v\lt0$: the particle moves back.

(c) Positions $x(0)=2\ \text{m}$, $x(1)=6\ \text{m}$, $x(3)=2\ \text{m}$, $x(4)=64-96+36+2=6\ \text{m}$. Displacement $=6-2=+4\ \text{m}$. Path length $=4+4+4=12\ \text{m}$ (split at the turning points!).

Averages $\bar v=\dfrac{4\ \text{m}}{4\ \text{s}}=1\ \text{m/s}$; average speed $=\dfrac{12\ \text{m}}{4\ \text{s}}=3\ \text{m/s}$.

Check At $t=2\ \text{s}$, $v=12-24+9=-3\ \text{m/s}$: negative, as expected between the turning points. Interactive 2 shows the cart reversing twice with an odometer.

The red chord joins the points at $t$ and $t+\Delta t$ on the x–t graph; its slope is the average velocity $\Delta x/\Delta t$. Press Play to shrink $\Delta t$ from $2\ \text{s}$ to $0.001\ \text{s}$ and watch the chord swing onto the blue tangent, whose slope is the instantaneous velocity $dx/dt$. Move $t$ to test other instants.

Motion:

$t$:2.0 s$\Delta t$:2 s

$\Delta t$—

$\Delta x$—

average velocity $\Delta x/\Delta t$—

instantaneous $v=dx/dt$—

Reading x–t graphs

Because velocity is the slope of the x–t graph, the shape of the graph tells the whole story of the motion at a glance.

Feature of the x–t graph	What the object is doing
horizontal line	at rest ($v=0$)
straight line sloping up / down	uniform motion in the $+x$ / $-x$ direction; steeper means faster
curve getting steeper	speeding up
curve getting flatter	slowing down
peak or valley (horizontal tangent)	momentarily at rest and turning back
crossing the $t$-axis	passing through the origin (says nothing about the velocity)

✏️ Example 5 — A delivery scooter

The x–t graph of a delivery scooter on a straight road is made of straight segments joining the points $(0\ \text{s},\,0\ \text{m})$, $(40\ \text{s},\,600\ \text{m})$, $(100\ \text{s},\,600\ \text{m})$ and $(160\ \text{s},\,0\ \text{m})$. (a) Find the velocity in each segment and describe the motion. (b) Find the average velocity and average speed for the whole $160\ \text{s}$.

💡 Show solution▶

0–40 s slope $=\dfrac{600}{40}=+15\ \text{m/s}$: out along the road at a steady $54\ \text{km/h}$.

40–100 s slope $0$: at rest for $60\ \text{s}$ (handing over the parcel).

100–160 s slope $=\dfrac{0-600}{60}=-10\ \text{m/s}$: back towards the start at $36\ \text{km/h}$.

(b) $\Delta x=0$, so $\bar v=0$. Average speed $=\dfrac{1200\ \text{m}}{160\ \text{s}}=7.5\ \text{m/s}$.

Note At the corners ($t=40$ and $100\ \text{s}$) the slope jumps, so the velocity is not defined there. A real scooter rounds these corners off — it needs a short time to speed up or slow down — which is the subject of Section 2.2.

⚠️ Common mistakes

1. Path length is not displacement. Split the motion at every turning point (where $v=0$) before adding distances; Example 4 gives $12\ \text{m}$, not $4\ \text{m}$.

2. Average speed is not the average of the speeds. Divide total path length by total time; equal distances at $40$ and $60\ \text{km/h}$ average $48\ \text{km/h}$, not $50$.

3. Velocity is the slope, not the height, of the x–t graph. A body at $x=500\ \text{m}$ may be at rest; a body at $x=0$ may be moving fast.

4. An x–t graph is not a picture of the path. The motion is along a straight line; the curve only shows how $x$ changes with $t$.

5. Impossible graphs. A real x–t graph cannot have a vertical segment (infinite speed), cannot go backwards in time, and cannot have two values of $x$ at one $t$.

The cart moves along the track while its x–t graph is drawn below. The odometer adds up every metre covered (path length); the displacement only compares where the cart is now with where it started. The blue tangent's slope is the velocity. Pick a motion and press Play, or drag the time slider.

Motion:time:

time $t$—

position $x$—

displacement $x-x_0$—

path length (odometer)—

velocity $v$—

Exercises

Position, displacement and path length. Unless told otherwise, the origin is at a bus stop and $+x$ points east.

1 A cyclist starts at $x=-120\ \text{m}$ and rides to $x=+380\ \text{m}$. Find her displacement.
2 She then rides back to $x=+80\ \text{m}$. Find the displacement and the path length for the whole ride.
3 Repeat part 2 with the origin moved to the old $x=+200\ \text{m}$. Which answers change?
4 A lift starts at a floor $8.0\ \text{m}$ above the ground floor, rises to $28.0\ \text{m}$, then comes down to the ground floor. Taking $+$ upwards from the ground floor, find its displacement and path length.
5 A swimmer swims four lengths of a $50\ \text{m}$ pool (there and back twice). Find the displacement and path length.
6 A ball dropped from $5.0\ \text{m}$ above the floor bounces up to $3.0\ \text{m}$ and is caught there. Taking $+$ upwards from the floor, find its displacement and path length.
7 Can the magnitude of a displacement ever exceed the path length? When are they equal?

💡 Show solutions▶

1 $\Delta x=380-(-120)=+500\ \text{m}$, i.e. $500\ \text{m}$ east.

2 $\Delta x=80-(-120)=+200\ \text{m}$ (east); path length $500+300=800\ \text{m}$.

3 Positions become $-320$, $+180$, $-120\ \text{m}$ (each minus $200$). Displacement $-120-(-320)=+200\ \text{m}$ and path length $800\ \text{m}$ are unchanged; only positions depend on the origin.

4 $\Delta x=0-8.0=-8.0\ \text{m}$ (down); path length $20.0+28.0=48.0\ \text{m}$.

5 Back at the start: $\Delta x=0$; path length $4\times50=200\ \text{m}$.

6 $\Delta x=3.0-5.0=-2.0\ \text{m}$ (down); path length $5.0+3.0=8.0\ \text{m}$.

7 Never: the straight-line gap between start and end cannot exceed the length of any path joining them. They are equal when the motion is in one direction without turning back.

Average velocity and average speed.

1 A train covers $150\ \text{km}$ along a straight track in $2.5\ \text{h}$. Find its average speed in $\text{km/h}$ and in $\text{m/s}$.
2 The cyclist of Card I takes $50\ \text{s}$ to ride from $-120\ \text{m}$ to $+380\ \text{m}$ and $40\ \text{s}$ to return to $+80\ \text{m}$. Find her average velocity and average speed for the whole ride.
3 A car covers $30\ \text{km}$ at $60\ \text{km/h}$ and the next $30\ \text{km}$, in the same direction, at $40\ \text{km/h}$. Find its average speed.
4 Another car drives for $1\ \text{h}$ at $60\ \text{km/h}$ and then for $1\ \text{h}$ at $40\ \text{km/h}$ in the same direction. Find its average speed, and explain why it differs from part 3.
5 A jogger runs $3.0\ \text{km}$ east in $15\ \text{min}$, then walks $1.0\ \text{km}$ west in $15\ \text{min}$. Find her average velocity and average speed in $\text{km/h}$.
6 Show that the magnitude of the average velocity over any interval is never greater than the average speed.
7 A boy walks to a shop at speed $v_1$ and returns along the same road at $v_2$. Show that his average speed is $\dfrac{2v_1v_2}{v_1+v_2}$, and evaluate it for $v_1=1.2\ \text{m/s}$, $v_2=1.8\ \text{m/s}$.

💡 Show solutions▶

1 $\dfrac{150}{2.5}=60\ \text{km/h}=\dfrac{60}{3.6}=16.7\ \text{m/s}$.

2 $\bar v=\dfrac{+200\ \text{m}}{90\ \text{s}}=+2.2\ \text{m/s}$ (east); average speed $=\dfrac{800\ \text{m}}{90\ \text{s}}=8.9\ \text{m/s}$.

3 Times $0.50\ \text{h}$ and $0.75\ \text{h}$: average speed $=\dfrac{60\ \text{km}}{1.25\ \text{h}}=48\ \text{km/h}$.

4 $\dfrac{60+40}{2}=50\ \text{km/h}$. With equal times the average is the simple mean; with equal distances (part 3) the car spends longer at the lower speed, which pulls the average down.

5 $\Delta x=+2.0\ \text{km}$ in $0.50\ \text{h}$: $\bar v=4.0\ \text{km/h}$ east; average speed $=\dfrac{4.0\ \text{km}}{0.50\ \text{h}}=8.0\ \text{km/h}$.

6 $|\Delta x|\le$ path length (Card I, part 7); dividing both by the same positive $\Delta t$ keeps the inequality.

7 For a one-way distance $d$: total time $\dfrac{d}{v_1}+\dfrac{d}{v_2}$, so average speed $=\dfrac{2d}{d/v_1+d/v_2}=\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2(1.2)(1.8)}{3.0}=1.44\approx1.4\ \text{m/s}$. (His average velocity is zero.)

III

Instantaneous velocity. Positions are in metres and times in seconds; use Result 2.1.

1 $x=5.0\,t^2$. Find $v$ at $t=3.0\ \text{s}$.
2 $x=3.0+4.0\,t$. Find $v$ at any time. What kind of motion is this?
3 $x=12t-3.0\,t^2$. Find $v(t)$, the time at which the body is momentarily at rest, and its position then.
4 $x=2.0\,t^3$. Find the average velocity between $t=1.0\ \text{s}$ and $t=3.0\ \text{s}$ and the instantaneous velocity at $t=2.0\ \text{s}$. Why are they different?
5 $x=t^2$. Find the average velocity over $[3,\,3.1]$ and $[3,\,3.01]$ and guess $v(3)$. Confirm with Result 2.1.
6 $x=40-5.0\,t^2$. Find the velocity and the speed at $t=2.0\ \text{s}$, and the direction of motion.
7 $x=t^3-9t^2+24t$. At what times is the particle momentarily at rest, and where is it then?
8 For the particle of part 7, find the displacement and path length between $t=0$ and $t=5\ \text{s}$.
9 In $x=At+Bt^3$, find the SI units of $A$ and $B$, and write $v(t)$. Check that every term of $v$ has the unit $\text{m/s}$.

💡 Show solutions▶

1 $v=10t$; $v(3.0)=30\ \text{m/s}$.

2 $v=4.0\ \text{m/s}$ at all times: uniform motion in the $+x$ direction, starting from $x=3.0\ \text{m}$.

3 $v=12-6.0\,t$; $v=0$ at $t=2.0\ \text{s}$, where $x=24-12=12\ \text{m}$ (the turning point).

4 $\bar v=\dfrac{54-2.0}{2.0}=26\ \text{m/s}$; $v=6.0\,t^2$ gives $v(2.0)=24\ \text{m/s}$. The x–t graph curves, so a chord's slope need not equal the tangent's slope at the midpoint time.

5 $\dfrac{9.61-9}{0.1}=6.1\ \text{m/s}$; $\dfrac{9.0601-9}{0.01}=6.01\ \text{m/s}$; guess $6\ \text{m/s}$. Result 2.1: $v=2t=6\ \text{m/s}$ ✓.

6 $v=-10t$; at $2.0\ \text{s}$, $v=-20\ \text{m/s}$, speed $20\ \text{m/s}$, moving in the $-x$ direction.

7 $v=3t^2-18t+24=3(t-2)(t-4)=0$ at $t=2\ \text{s}$ ($x=8-36+48=20\ \text{m}$) and $t=4\ \text{s}$ ($x=64-144+96=16\ \text{m}$).

8 $x(0)=0$, $x(2)=20$, $x(4)=16$, $x(5)=125-225+120=20\ \text{m}$. Displacement $+20\ \text{m}$; path length $20+4+4=28\ \text{m}$.

9 $[A]=[\text{L}\,\text{T}^{-1}]$, unit $\text{m/s}$; $[B]=[\text{L}\,\text{T}^{-3}]$, unit $\text{m/s}^3$. $v=A+3Bt^2$: $A$ in $\text{m/s}$, and $Bt^2$ in $(\text{m/s}^3)(\text{s}^2)=\text{m/s}$ ✓.

x–t graphs and applications. In parts 1–4 the x–t graph is made of straight segments joining $(0\ \text{s},0\ \text{m})$, $(10\ \text{s},50\ \text{m})$, $(20\ \text{s},50\ \text{m})$, $(30\ \text{s},150\ \text{m})$ and $(50\ \text{s},0\ \text{m})$.

1 Find the velocity in each of the four segments.
2 When is the body at rest? In which segment is it fastest?
3 Find the displacement and path length over the whole $50\ \text{s}$.
4 Find the average velocity and average speed over the whole $50\ \text{s}$, and the average velocity from $t=0$ to $t=30\ \text{s}$.
5 Explain why an x–t graph can never contain a vertical segment.
6 A car moves with $x=0.80\,t^2$ for $0\le t\le10\ \text{s}$. Find its average velocity over these $10\ \text{s}$ and the instant at which its velocity equals that average.
7 A phone logs a cyclist's position every $2.0\ \text{s}$: $t=0,2,4,6,8\ \text{s}$; $x=0,\ 9.0,\ 20.0,\ 33.0,\ 48.0\ \text{m}$. Estimate her velocity at $t=4.0\ \text{s}$ and at $t=6.0\ \text{s}$ using the average over the interval centred on each instant.
8 (Modelling) Two cameras $10.5\ \text{km}$ apart on an expressway with a $100\ \text{km/h}$ limit record a car at 10:02:10 and 10:07:55. Find its average speed. Must it have been over the limit at some instant? Explain.

💡 Show solutions▶

1 $\dfrac{50}{10}=5.0\ \text{m/s}$; $\ 0$; $\ \dfrac{100}{10}=10\ \text{m/s}$; $\ \dfrac{0-150}{20}=-7.5\ \text{m/s}$.

2 At rest from $10$ to $20\ \text{s}$. Fastest from $20$ to $30\ \text{s}$ (speed $10\ \text{m/s}$; the last segment has speed only $7.5\ \text{m/s}$).

3 Displacement $0$ (back at the origin); path length $50+0+100+150=300\ \text{m}$.

4 $\bar v=0$; average speed $=\dfrac{300}{50}=6.0\ \text{m/s}$; from $0$ to $30\ \text{s}$, $\bar v=\dfrac{150}{30}=5.0\ \text{m/s}$.

5 A vertical segment would mean a change of position in zero time — an infinite velocity — and several positions at one instant. Neither is possible.

6 $\bar v=\dfrac{80-0}{10}=8.0\ \text{m/s}$; $v=1.6\,t=8.0$ at $t=5.0\ \text{s}$.

7 $v(4.0)\approx\dfrac{33.0-9.0}{4.0}=6.0\ \text{m/s}$; $v(6.0)\approx\dfrac{48.0-20.0}{4.0}=7.0\ \text{m/s}$. (The data fit $x=4t+0.25t^2$, for which these estimates are exact.)

8 $\Delta t=5\ \text{min}\ 45\ \text{s}=345\ \text{s}$; average speed $=\dfrac{10\,500\ \text{m}}{345\ \text{s}}=30.4\ \text{m/s}=110\ \text{km/h}$. Yes: if the speed had never exceeded $100\ \text{km/h}$, the car could have covered at most $100\ \text{km/h}\times345\ \text{s}=9.58\ \text{km}$. An average above the limit forces at least one instant above it.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Distinguish between displacement and path length. Give an example in which the path length is non-zero but the displacement is zero.

💡 Show solution▶

Difference Displacement $x_2-x_1$ is the change of position, with a sign (direction), depending only on the end points; path length is the total distance actually covered, never negative, depending on the whole path. [1]

Example A swimmer completing one length of a $50\ \text{m}$ pool and back: path length $100\ \text{m}$, displacement $0$. [1]

Exam tip Mention that $\text{path length}\ge|\text{displacement}|$ — it shows understanding and often earns the full mark.

E2. NEET A body goes from town A to town B at $20\ \text{m/s}$ and returns along the same road at $30\ \text{m/s}$. Its average velocity and average speed for the round trip are (a) $0$ and $24\ \text{m/s}$ (b) $25\ \text{m/s}$ and $25\ \text{m/s}$ (c) $0$ and $25\ \text{m/s}$ (d) $24\ \text{m/s}$ and $0$

💡 Show solution▶

Velocity Back at the start: $\Delta x=0$, so $\bar v=0$.

Speed Equal distances: $\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2(20)(30)}{50}=24\ \text{m/s}$.

Answer (a).

Exam tip (c) is the trap: $25$ is the simple mean, valid only for equal times. The harmonic mean is always below the simple mean.

E3. NEET JEE Main The position of a particle is $x=4.0+6.0\,t-0.50\,t^3$ ($x$ in m, $t$ in s). At $t=2.0\ \text{s}$ its velocity is (a) $0$ (b) $6.0\ \text{m/s}$ (c) $12\ \text{m/s}$ (d) $-6.0\ \text{m/s}$

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Differentiate $v=\dfrac{dx}{dt}=6.0-1.5\,t^2$.

Substitute $v(2.0)=6.0-1.5(4.0)=0$.

Answer (a) — the particle is momentarily at rest at $x=4.0+12-4.0=12\ \text{m}$.

Exam tip (b) forgets the $t^3$ term; (c) is the position. Never confuse $x$ with $dx/dt$.

E4. JEE Main (Numerical value) A particle moves along the $x$-axis with $x=2t^3-15t^2+36t$ ($x$ in m, $t$ in s). Find the distance (path length) it covers between $t=0$ and $t=4\ \text{s}$.

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Turning points $v=6t^2-30t+36=6(t-2)(t-3)=0$ at $t=2$ and $3\ \text{s}$.

Positions $x(0)=0$, $x(2)=16-60+72=28\ \text{m}$, $x(3)=54-135+108=27\ \text{m}$, $x(4)=128-240+144=32\ \text{m}$.

Answer Path length $=28+1+5=34\ \text{m}$ (the displacement is only $32\ \text{m}$).

Exam tip Always find where $v=0$ inside the interval first; the answer $32$ is the commonest wrong choice.

E5. JEE Advanced (One or more correct) A particle moves with $x=10+8t-t^2$ ($x$ in m, $t$ in s) for $0\le t\le8\ \text{s}$. Which statements are correct? (A) It is momentarily at rest at $t=4\ \text{s}$. (B) Its average velocity over the $8\ \text{s}$ is zero. (C) The path length over the $8\ \text{s}$ is $32\ \text{m}$. (D) Its average speed over the $8\ \text{s}$ is $2\ \text{m/s}$.

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(A) $v=8-2t=0$ at $t=4\ \text{s}$ ✓.

(B) $x(0)=10\ \text{m}$ and $x(8)=10+64-64=10\ \text{m}$: $\Delta x=0$, so $\bar v=0$ ✓.

(D) Average speed $=\dfrac{32}{8}=4\ \text{m/s}$, not $2$ ✗.

Answer (A), (B), (C). Exam tip: (D) halves the path length — a slip that comes from using only the outward leg.

E6. Board · 3 marks A car travels the first half of a straight journey at $40\ \text{km/h}$ and the second half at $60\ \text{km/h}$. (a) Find its average speed. (b) Find its average speed if instead it travels for the first half of the time at $40\ \text{km/h}$ and the second half at $60\ \text{km/h}$. (c) Why are the answers different?

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(a) For total distance $2d$: time $=\dfrac d{40}+\dfrac d{60}=\dfrac{d}{24}$, so average speed $=\dfrac{2d}{d/24}=48\ \text{km/h}$. [1]

(b) For total time $2T$: distance $=40T+60T=100T$, average speed $=\dfrac{100T}{2T}=50\ \text{km/h}$. [1]

(c) In (a) the car spends more time at the lower speed ($1.5$ times as long), so the average is pulled below the simple mean. [1]

Exam tip Always go back to the definition, total path length ÷ total time, rather than averaging speeds.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Can an average-speed camera prove that a driver broke the speed limit at some instant — and how do road authorities choose the distance between the two cameras?

Why it matters. Point-to-point ("section control") speed enforcement is used on motorways in the United Kingdom, the Netherlands, Austria, Australia and on several Indian expressways. Instead of catching a car at one spot, it times the car between two cameras kilometres apart. Road-safety studies report that it smooths traffic and reduces serious crashes along the whole section, not just near the camera. Its legal basis is exactly the link between average and instantaneous velocity in this section.

What to do.

Explain, using an x–t graph, why a car whose average speed over a section exceeds the limit must have exceeded the limit at some instant (compare Card IV, part 8).
Is the converse true? Sketch an x–t graph for a car that exceeds the limit briefly but still averages below it.
Clocks at the two cameras are synchronised to within $\pm0.1\ \text{s}$. For sections of $0.5\ \text{km}$, $2\ \text{km}$ and $10\ \text{km}$, find the uncertainty in a measured average speed near $100\ \text{km/h}$. Which spacing gives the fairest fine?
Record a short journey (as a passenger) with a phone GPS-logging app. Compare the speed it shows at a few instants with the average over the whole trip.

Tools: Graph paper or a spreadsheet; a phone with a free GPS logger (optional).

Hand in: Your two sketches, a table of timing uncertainties for the three section lengths, and a paragraph recommending a section length with reasons.

Going further. The statement in step 1 is a version of the mean value theorem, which you will meet in Class 12 Mathematics. Find out what condition on the motion it needs.

Instructor note: Near $100\ \text{km/h}=27.8\ \text{m/s}$, a $0.5\ \text{km}$ section takes $18\ \text{s}$, so a $0.1\ \text{s}$ timing error is $0.56\%$ ($\approx0.6\ \text{km/h}$); at $2\ \text{km}$ ($72\ \text{s}$) it is $0.14\%$ ($0.14\ \text{km/h}$); at $10\ \text{km}$ ($360\ \text{s}$) $0.028\%$ ($0.03\ \text{km/h}$). Longer sections are more precise but say less about where the speeding happened. The converse in step 2 is false — the reason point cameras are still used at hazards.

🧪 Self-check quiz

0/5 correct

Q1. A runner completes exactly two laps of a straight out-and-back course of length $200\ \text{m}$ each way, finishing at the start. Her displacement and path length are

Q2. On an x–t graph, the instantaneous velocity at time $t$ is

Q3. For $x=3t^2-2t+5$ (SI units), the velocity at $t=2\ \text{s}$ is

Q4. Which of the following can never be the x–t graph of a real body?

Q5. A car covers equal distances at $30\ \text{km/h}$ and $60\ \text{km/h}$ in the same direction. Its average speed is

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