NDA Physics · Kinematics and Motion

Equations of Motion and Motion Graphs

For motion with constant acceleration the three equations v = u + at, s = ut + ½at², and v² = u² + 2as link the five quantities u, v, a, s, t; a motion graph reads the same physics off slopes (acceleration) and areas (displacement).

All Kinematics and Motion notes NDA Physics strategy

Why this matters

This is the engine room of the chapter — 15 PYQs, the most of any subtopic, and the source of most of its HARD questions. Almost every numerical reduces to picking the right one of the three equations and substituting carefully (watch the sign of a when decelerating). The graph questions test two rules over and over: on a velocity-time graph the slope is the acceleration and the area is the displacement; on a position-time graph the slope is the velocity. Master the substitution discipline and the graph-reading rules and this subtopic is yours.

Concept 1 of 6

Acceleration — rate of change of velocity

Intuition

Acceleration measures how fast the velocity is changing. Speeding up gives positive acceleration; slowing down (deceleration) gives negative acceleration in the direction of motion. A change in direction at constant speed is still an acceleration, because velocity is a vector.

Definition

Acceleration is the rate of change of velocity: $a = \dfrac{v - u}{t}$ (a vector, SI unit m/s $^2$ ). Uniform (constant) acceleration means $a$ does not change with time. A negative value in the direction of motion is a deceleration (retardation). Even with constant speed, turning is an acceleration.

Acceleration

a = \dfrac{v - u}{t}

aacceleration (m/s²)
uinitial velocity
vfinal velocity
ttime interval

Worked example

A car's velocity rises from 10 m/s to 30 m/s in 5 s. Find its acceleration.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Kinematics and MotionMODERATE

Consider the following velocity and time graph. [Graph shows velocity = 8 m/s from t=0 to t=8, then drops to 4 m/s at t=8 and stays constant till t=12, then drops at t=12.] Which one of the following is the value of average acceleration from 8 s to 12 s?

[Q109 · Sep · 2018]

Deceleration is negative acceleration, not 'no' acceleration

Slowing down is still acceleration — just opposite to the velocity, so it carries a minus sign in the direction of motion. From a velocity-time graph, average acceleration over an interval = (change in velocity) / (time), e.g. (4 − 8)/(12 − 8) = −1 m/s².

Drill 1 more on acceleration — rate of change of velocity

Concept 2 of 6

The three equations of motion

Intuition

Three equations connect the five quantities u, v, a, s, t for constant acceleration. Each equation leaves out exactly one of the five — pick the one missing the quantity you neither know nor want. They only hold when acceleration is constant.

Definition

For constant acceleration:

$v = u + at$ (no $s$ ) — velocity after time $t$ .
$s = ut + \tfrac{1}{2}at^2$ (no $v$ ) — displacement after time $t$ .
$v^2 = u^2 + 2as$ (no $t$ ) — velocity after displacement $s$ .

Use a consistent sign convention: take one direction as positive and give $a$ a minus sign when it opposes the motion.

Equations of motion (constant a)

v = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as

uinitial velocity
vfinal velocity
aacceleration (constant)
sdisplacement
ttime

Worked example

A car at 12 m/s brakes uniformly and stops in 45 m. Find its acceleration.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Kinematics and MotionMODERATE

A car has an initial velocity of 12 m/s and is brought to rest over a distance of 45 m. The acceleration of the car is.

[Q130 · Apr · 2025]

v² − u² = 2as, with the right sign

The third equation is

v^2 - u^2 = 2as

, i.e.

v^2 = u^2 + 2as

. A common wrong form writes

u^2 - v^2 = 2as

— that flips the sign and is the false option the bank tests. When decelerating, keep

a

negative rather than reordering

u

and

v

These equations need CONSTANT acceleration

v = u + at and friends apply only while a is constant. For a journey in two phases with different accelerations, apply the equations to each phase separately and add the results (as in the 'first t s at 2 m/s², next 10 s at 5 m/s²' problem).

Drill 3 more on the three equations of motion

Concept 3 of 6

Distance covered in the nth second

Intuition

The distance covered during one particular second (say the 5th second) is not the same as the total distance after 5 seconds. It is the gap between the total after n seconds and the total after (n−1) seconds, which simplifies to a neat formula.

Definition

The distance travelled during the nth second of uniformly accelerated motion is $s_n = u + \tfrac{1}{2}a(2n - 1)$ . It is a distance covered in a 1-second interval, derived as $s_{(n)} - s_{(n-1)}$ from $s = ut + \tfrac{1}{2}at^2$ .

Distance in the nth second

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

s_ndistance during the nth second
uinitial velocity
aacceleration
nthe second of interest

Worked example

A body starts from rest with acceleration 4 m/s². How far does it travel during the 3rd second?

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 3Kinematics and MotionMODERATE

Which one of the following equations related to the motion of an object is NOT correct? (Symbols carry their usual meanings)

[Q145 · Apr · 2025]

The nth-second distance is not the total distance

sₙ = u + ½a(2n−1) gives the distance in a 1-second slice, not the cumulative s = ut + ½at². The valid NDA form is u + ½a(2n−1); options that drop the ½ or write (2n+1) are wrong.

Concept 4 of 6

Reading a velocity-time graph

Intuition

On a velocity-time graph, how steep the line is tells you the acceleration, and how much area sits under the line tells you the displacement. Uniform acceleration plots as a straight line; uniform velocity plots as a flat horizontal line.

Definition

On a velocity-time graph:

the slope = acceleration (a straight line ⟹ uniform acceleration; a horizontal line ⟹ zero acceleration, constant velocity);
the area between the line and the time axis = displacement.

An upward-sloping segment is acceleration; a downward-sloping segment is deceleration.

Velocity-time graph readings

a = \text{slope} = \dfrac{\Delta v}{\Delta t}, \qquad s = \text{area under the graph}

$\Delta v$ change in velocity
$\Delta t$ time interval
sdisplacement

On a velocity-time graph the slope of the line is the acceleration, and the area under the line is the displacement.

Worked example

A velocity-time graph is a straight line rising from 4 m/s at t = 0 to 12 m/s at t = 4 s. Find the acceleration and the displacement in those 4 s.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Kinematics and MotionMODERATE

[Q109 · Sep · 2018]

Slope is acceleration; AREA is displacement — don't swap them

On a velocity-time graph the slope gives acceleration and the area gives displacement. A frequent error reads the area as a velocity or the slope as a distance. Also: a segment with positive slope (CD) is the accelerated part, a negative-slope segment (AB) is decelerated.

Drill 3 more on reading a velocity-time graph

Concept 5 of 6

Reading a position-time graph

Intuition

On a position-time (x-t) graph the slope is the velocity. A straight line means constant velocity; a curve that gets steeper means the velocity is rising — acceleration. Watch the axis order: if TIME is on the vertical axis and position on the horizontal, the speed is 1/slope, so a steeper line means a SLOWER object.

Definition

On a position-time graph (x vertical, t horizontal): the slope $= \dfrac{dx}{dt} =$ velocity. A straight line ⟹ constant velocity; a curve ⟹ changing velocity (acceleration). If the axes are swapped so that time is plotted against position (t vertical, x horizontal), then speed $= dx/dt = 1/\text{slope}$ : a steeper t-x line means a lower speed.

Position-time slope

v = \dfrac{dx}{dt} = \text{slope of the } x\text{-}t \text{ graph}

xposition
ttime
vvelocity (slope)

On a position-time graph the slope is the velocity. A straight line means constant velocity; a steepening curve means the velocity is rising, i.e. acceleration.

Worked example

A position-time graph is a straight line from (0 s, 0 m) to (4 s, 20 m). What is the velocity, and is the motion accelerated?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Kinematics and MotionMODERATE

The figure shown above gives the time (

t

) versus position (

x

) graphs of three objects

A

B

and

C

. Which one of the following is the correct relation between their speeds

V_A

V_B

and

V_C

, respectively at any instant (

t > 0

[Q112 · Apr · 2019]

Check which axis is which before reading the slope

On the usual x-t graph the slope is the velocity directly. But if the figure plots TIME on the vertical axis against position, then speed = 1/slope, so the steepest line is the SLOWEST object — the reverse of the instinct.

Concept 6 of 6

Interpreting motion: shapes and statements

Intuition

Beyond reading numbers off a graph, the NDA asks you to match a physical situation to the right graph shape, or to judge which statement about a motion is true. A skydiver speeds up, levels off at terminal velocity, then slows when the parachute opens — a smooth rounded curve, not a sharp triangle.

Definition

Key interpretation rules:

For uniform acceleration from rest, distance grows quadratically with time (a parabola), not linearly.
With constant non-zero acceleration, the distance covered depends on the initial velocity u and the time, not on any initial displacement.
A skydiver accelerates, approaches terminal velocity (curve flattens), then decelerates after the parachute opens — a smooth rounded rise and gradual fall.
Treating $v = u + at$ as a distance-time relation gives a straight line with a positive intercept when $u \neq 0$ .

Worked example

An object moves with constant non-zero acceleration for a fixed time. Does the distance it covers depend on its initial velocity?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6Kinematics and MotionMODERATE

If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time

[Q63 · Sep · 2019]

Quadratic, not linear, distance growth

Under constant acceleration the distance grows as t² (a parabola), so 'distance increases linearly with time' is false. And it depends on the initial velocity u, not on any starting position.

Drill 2 more on interpreting motion: shapes and statements

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 (5)

Acceleration — rate of change of velocity
Acceleration
$a = \dfrac{v - u}{t}$
The three equations of motion
Equations of motion (constant a)
$v = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as$
Distance covered in the nth second
Distance in the nth second
$s_n = u + \tfrac{1}{2}a(2n - 1)$
Reading a velocity-time graph
Velocity-time graph readings
$a = \text{slope} = \dfrac{\Delta v}{\Delta t}, \qquad s = \text{area under the graph}$
Reading a position-time graph
Position-time slope
$v = \dfrac{dx}{dt} = \text{slope of the } x\text{-}t \text{ graph}$

Watch out for (7)

Deceleration is negative acceleration, not 'no' acceleration
→ Acceleration — rate of change of velocity
v² − u² = 2as, with the right sign
→ The three equations of motion
These equations need CONSTANT acceleration
→ The three equations of motion
The nth-second distance is not the total distance
→ Distance covered in the nth second
Slope is acceleration; AREA is displacement — don't swap them
→ Reading a velocity-time graph
Check which axis is which before reading the slope
→ Reading a position-time graph
Quadratic, not linear, distance growth
→ Interpreting motion: shapes and statements

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