CDS Physics · Kinematics and Motion

Motion in a Straight Line

Distance and displacement, average speed, the three equations of uniform acceleration, and what the slope and the area of a motion graph tell you.

All Kinematics and Motion notes

Why this matters

Ten CDS questions from 2018 (II) to 2025 (II), nine of them MODERATE. Four are short sums on the equations of motion, three ask for a displacement or an average speed, and three read a motion graph.

Concept 1 of 3: Distance, displacement and average speed

Distance is the length of the path you actually walk. Displacement is the straight-line gap from where you started to where you stopped, with a direction. Run once round a track and you have covered a lot of distance, but your displacement is zero, because you are back at the start. Average speed is always total distance divided by total time, never the plain average of two speeds.

Definition

The rules:

Distance is a scalar and never decreases. Displacement is a vector: final position minus starting position.
On a circular track, after any whole number of laps the displacement is zero; after half a lap (or 1.5, 2.5 laps and so on) it equals the diameter.
Average speed = total distance ÷ total time.
Two equal distances at speeds $v_1$ and $v_2$ : average speed $= \dfrac{2v_1v_2}{v_1 + v_2}$ , which is always less than the plain average.
Two equal times at speeds $v_1$ and $v_2$ : average speed $= \dfrac{v_1 + v_2}{2}$ .

Average speed

\bar{v} = \dfrac{\text{total distance}}{\text{total time}} \qquad \bar{v}_{\text{equal distances}} = \dfrac{2v_1v_2}{v_1 + v_2} \qquad \bar{v}_{\text{equal times}} = \dfrac{v_1 + v_2}{2}

v₁, v₂the speeds on the two parts of the trip

Worked example

A cyclist goes round a circular track of radius 35 m, taking 40 s for each lap. Find the distance covered and the displacement after 100 s. (Take

\pi = 22/7

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (I) 2022 — General Knowledge · Q5Moderate

Example 1 · Kinematics and Motion · Motion in a Straight Line

A boy completes one round of a circular track of diameter 200 m in 30 s. What will be the displacement at the end of 3 minutes and 45 seconds ?

Half a lap gives the diameter, not half the circumference

Displacement is the straight line between start and finish. After half a lap that line is the diameter. Half the circumference is the distance walked, which is a different answer.

Average speed over equal distances is not the plain average

Over equal distances the slower part takes longer, so the average is pulled down:

\dfrac{2v_1v_2}{v_1 + v_2}

. The plain average

\dfrac{v_1 + v_2}{2}

is right only for equal times.

Count whole laps first

Divide the time by the time for one lap. A whole number of laps means zero displacement; a half lap left over means one diameter. Do this before any other working.

Drill 2 more on distance, displacement and average speed

Concept 2 of 3: Equations of motion for uniform acceleration

When a body speeds up or slows down at a steady rate, three short equations link its starting velocity, final velocity, acceleration, time and displacement. Pick the one that leaves out the quantity you were not given. A fourth shortcut is often quickest: with steady acceleration the average velocity is halfway between the start and the end, so displacement is that average times the time.

Definition

For constant acceleration $a$ , starting velocity $u$ , final velocity $v$ , time $t$ and displacement $s$ :

$v = u + at$ (no $s$ )
$s = ut + \tfrac{1}{2}at^{2}$ (no $v$ )
$v^{2} = u^{2} + 2as$ (no $t$ ); for braking to rest, $s = \dfrac{u^{2}}{2a}$ with $a$ the retardation.
$s = \dfrac{u + v}{2}\,t$ (no $a$ )
Distance in the n-th second: $s_n = u + \tfrac{a}{2}(2n - 1)$ . From rest, the distances in successive seconds go 1 : 3 : 5 : 7 : ...
Keep units matched: km/h with hours, m/s with seconds.

Equations of motion

v = u + at \qquad s = ut + \tfrac{1}{2}at^{2} \qquad v^{2} = u^{2} + 2as \qquad s = \dfrac{u + v}{2}\,t \qquad s_n = u + \dfrac{a}{2}(2n - 1)

u, vstarting and final velocity
aconstant acceleration (negative when slowing down)
sdisplacement in time t
sₙdisplacement in the n-th second alone

Worked example

A train running at 20 m/s brakes with a steady retardation of 4 m/s². How long does it take to stop, and how far does it go in that time?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (I) 2023 — General Knowledge · Q48Moderate

Example 2 · Kinematics and Motion · Motion in a Straight Line

A car moving with a speed of 12 m/s is subjected to brakes which produces a deceleration of

6\ \text{m/s}^2

. The car takes 2 s to stop after the application of brakes. What is the distance covered by the car after the application of brakes?

Change minutes into hours before dividing

If the speed is in km/h, the time must be in hours. Ten minutes is

\tfrac{1}{6}

h, not 10. Mixing the two gives an answer off by a factor of 60.

The n-th second is one second, not n seconds

The distance in the n-th second is

u + \tfrac{a}{2}(2n - 1)

, the gap between the distance after n seconds and after (n − 1) seconds. Using

\tfrac{1}{2}an^{2}

gives the total from the start instead.

Retardation enters with a minus sign

When a body slows down, put

a

in as a negative number in

v = u + at

and

v^{2} = u^{2} + 2as

, or use the stopping form

s = \dfrac{u^{2}}{2a}

with the retardation as a positive number. Do not do both.

Drill 3 more on equations of motion for uniform acceleration

Concept 3 of 3: Reading motion graphs

A motion graph puts time along the bottom. On a distance-time graph the steepness of the line is the speed: a flat line means the body is not moving, a steeper line means it is moving faster. On a velocity-time graph the steepness is the acceleration, and the area under the line is the displacement.

Definition

Two graphs, two readings each:

Distance-time (or position-time) graph: slope = speed (velocity). Horizontal line: at rest. Straight sloping line: uniform speed. Curve getting steeper: speeding up.
Velocity-time graph: slope = acceleration; area under the graph = displacement. Area below the time axis counts as negative.
A horizontal velocity-time line means constant velocity (zero acceleration), not rest.
A position-time graph can never be a vertical line: that would mean being in two places at one instant.

Graph and shape	What it means
Distance-time, horizontal line	The body is at restQ CDS 2019 (II): for a body at rest, the time-distance graph is horizontal.
Distance-time, steeper part	Higher speed in that partQ CDS 2023 (II): three parts of one journey are ranked by the slope of each part.
Velocity-time, horizontal line	Constant velocity, zero acceleration
Velocity-time, slope	Acceleration
Velocity-time, area under the line	DisplacementQ CDS 2023 (I): the area under a velocity-time graph gives the net displacement.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (II) 2019 — General Knowledge · Q94Moderate

Example 3 · Kinematics and Motion · Motion in a Straight Line

If an object is at rest, then the time (X-axis) versus distance (Y-axis) graph

Flat on one graph is not flat on the other

A horizontal distance-time line means rest. A horizontal velocity-time line means constant velocity. Check which quantity is on the vertical axis before you read the shape.

Area gives displacement, not average velocity

The area under a velocity-time graph is the displacement. Divide it by the time if you want the average velocity. With part of the graph below the axis, the area gives the net displacement, not the total distance.

Drill 2 more on reading motion graphs

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Distance, displacement and average speed
Average speed
$\bar{v} = \dfrac{\text{total distance}}{\text{total time}} \qquad \bar{v}_{\text{equal distances}} = \dfrac{2v_1v_2}{v_1 + v_2} \qquad \bar{v}_{\text{equal times}} = \dfrac{v_1 + v_2}{2}$
Equations of motion for uniform acceleration
Equations of motion
$v = u + at \qquad s = ut + \tfrac{1}{2}at^{2} \qquad v^{2} = u^{2} + 2as \qquad s = \dfrac{u + v}{2}\,t \qquad s_n = u + \dfrac{a}{2}(2n - 1)$

Reference tables (1)

Reading motion graphs5 rows

Graph and shape	What it means
Distance-time, horizontal line	The body is at restQ CDS 2019 (II): for a body at rest, the time-distance graph is horizontal.
Distance-time, steeper part	Higher speed in that partQ CDS 2023 (II): three parts of one journey are ranked by the slope of each part.
Velocity-time, horizontal line	Constant velocity, zero acceleration
Velocity-time, slope	Acceleration
Velocity-time, area under the line	DisplacementQ CDS 2023 (I): the area under a velocity-time graph gives the net displacement.

Watch out for (8)

Half a lap gives the diameter, not half the circumference
→ Distance, displacement and average speed
Average speed over equal distances is not the plain average
→ Distance, displacement and average speed
Count whole laps first
→ Distance, displacement and average speed
Change minutes into hours before dividing
→ Equations of motion for uniform acceleration
The n-th second is one second, not n seconds
→ Equations of motion for uniform acceleration
Retardation enters with a minus sign
→ Equations of motion for uniform acceleration
Flat on one graph is not flat on the other
→ Reading motion graphs
Area gives displacement, not average velocity
→ Reading motion graphs

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