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.

$a = \dfrac{v - u}{t} = \dfrac{30 - 10}{5}$ .
$a = \dfrac{20}{5} = 4$ m/s $^2$ .

Answer:4 m/s².

Practice this conceptself-check · 4 quick reps

Try it yourself

A train slows from 20 m/s to 8 m/s in 6 s. Find its acceleration and say what the sign means.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Velocity 0 → 12 m/s in 4 s. Acceleration?
2.
Velocity 15 m/s → 0 in 3 s. Acceleration?
3.
SI unit of acceleration?
4.
Can a body moving at constant speed have acceleration?

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.

Known: $u = 12$ , $v = 0$ , $s = 45$ ; unknown $a$ ; $t$ not wanted → use $v^2 = u^2 + 2as$ .
$0 = 12^2 + 2a(45) = 144 + 90a$ .
$a = \dfrac{-144}{90} = -1.6$ m/s $^2$ (the minus sign = braking).

Answer:−1.6 m/s².

Practice this conceptself-check · 4 quick reps

Try it yourself

A body starts from rest with acceleration 2 m/s². How far does it travel in the first 10 s?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
From rest, a = 3 m/s². Velocity after 4 s?
2.
u = 5 m/s, a = 2 m/s². Distance in 3 s?
3.
u = 0, a = 4 m/s², s = 8 m. Final speed?
4.
Which equation has no time t?

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?

Use $s_n = u + \tfrac{1}{2}a(2n - 1)$ with $u = 0$ , $a = 4$ , $n = 3$ .
$s_3 = 0 + \tfrac{1}{2}(4)(2\cdot3 - 1) = 2(5) = 10$ m.

Answer:10 m.

Practice this conceptself-check · 3 quick reps

Try it yourself

A car moving at 6 m/s accelerates at 2 m/s². Distance covered in the 4th second?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
From rest, a = 2 m/s². Distance in the 1st second?
2.
From rest, a = 10 m/s². Distance in the 2nd second?
3.
Is sₙ a distance for one second or the total after n seconds?

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 vchange in velocity
\Delta ttime 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.

Acceleration = slope = $\dfrac{12 - 4}{4 - 0} = 2$ m/s $^2$ .
Displacement = area = trapezium = $\tfrac{1}{2}(4 + 12)(4) = \tfrac{1}{2}(16)(4) = 32$ m.

Answer:Acceleration 2 m/s²; displacement 32 m.

Practice this conceptself-check · 4 quick reps

Try it yourself

On a velocity-time graph the velocity falls from 8 m/s at t = 8 s to 4 m/s at t = 12 s. Find the average acceleration over that interval.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
A horizontal velocity-time line means what acceleration?
2.
Velocity-time slope gives which quantity?
3.
Area under a velocity-time graph gives which quantity?
4.
Rectangle of v = 5 m/s held for 6 s gives what displacement?

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?

Velocity = slope = $\dfrac{20 - 0}{4 - 0} = 5$ m/s.
The graph is straight, so the slope (velocity) is constant — no acceleration.

Answer:Velocity 5 m/s; uniform motion (no acceleration).

Practice this conceptself-check · 4 quick reps

Try it yourself

Three objects A, B, C are shown on a time-versus-position graph (t on the vertical axis). C's line is the steepest, A's the least steep. Rank their speeds.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
On an x-t graph, the slope is which quantity?
2.
A straight x-t line means what kind of motion?
3.
A curving x-t graph indicates what?
4.
On a t-x graph, a steeper line means a faster or slower object?

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?

Distance is $s = ut + \tfrac{1}{2}at^2$ .
The first term $ut$ carries the initial velocity u, so a larger u gives a larger s for the same a and t.
Hence the distance DOES depend on the initial velocity (and it grows quadratically, not linearly, in t).

Answer:Yes — distance depends on the initial velocity u (via the ut term).

Practice this conceptself-check · 4 quick reps

Try it yourself

Which speed-time shape matches a skydiver's jump: (a) a sharp rise then sharp drop, or (b) a rounded rise to a flat top then a gradual fall to zero?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
From rest with constant a, distance grows how with time?
2.
Does distance covered depend on initial DISPLACEMENT?
3.
v = u + at plotted as distance-time (u ≠ 0) is what shape?
4.
A skydiver's speed-time curve at terminal velocity does what?

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

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Kinematics and MotionMODERATE

The speed of a car travelling on a straight road is listed below at successive intervals of 1 s : Time (s): 0 1 2 3 4 Speed (m/s): 0 2 4 6 8 Which of the following is/are correct ? The car travels 1. with a uniform acceleration of 2 m/

s^{2}

. 2. 16 m in 4 s. 3. with an average speed of 4 m/s. Select the correct answer using the code given below :

[Q54 · Apr · 2017]

Example 2Kinematics and MotionHARD

A ball is released from rest and rolls down an inclined plane, as shown in the following figure, requiring 4 s to cover a distance of 100 cm along the plane. Which one of the following is the correct value of angle

\theta

that the plane makes with the horizontal? (

g = 1000

cm/s

^2

)

[Q111 · Sep · 2018]

Example 3Kinematics and MotionEASY

What is the nature of velocity-time graph for a car moving with uniform acceleration?

[Q97 · Sep · 2022]

Example 4Kinematics and MotionHARD

Which one of the following graphs represents the equation of motion

v = u + at

; where all quantities are non-zero and symbols carry their usual meanings?

[Q131 · Sep · 2023]

Example 5Kinematics 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]

Drill every past-year question on this subtopic

13 questions from the bank — paginated, with cart and Word-export support.

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