NDA Physics · Kinematics and Motion

Foundations: Vectors, Distance, Displacement, and Position

Motion is described with scalars (magnitude only — distance, speed) and vectors (magnitude and direction — displacement, velocity, acceleration); the position vector r(t) packages where a particle is at every instant.

All Kinematics and Motion notes NDA Physics strategy

Why this matters

Start here. Almost every wrong answer in this chapter comes from confusing a quantity with its magnitude: distance with displacement, speed with velocity, the position vector with its length. Three PYQs sit directly here — the scalar/vector classification, the position-vector r(t) (HARD), and a two-leg net-displacement problem (HARD) — and the distinction underpins the round-trip and average-velocity questions in the next subtopic. Get the definitions watertight and the rest of kinematics becomes arithmetic.

Concept 1 of 4

Scalars vs vectors

Intuition

A scalar is fully described by a number and a unit — how much. A vector also needs a direction — how much and which way. Distance is a scalar (5 km); displacement is a vector (5 km north). Speed is a scalar (30 m/s); velocity is a vector (30 m/s east). This single split decides most of the chapter's true/false options.

Definition

A scalar has magnitude only: distance, speed, time, mass, energy, temperature.
A vector has magnitude AND direction: displacement, velocity, acceleration, force, momentum.

Two vectors are equal only if both their magnitude and direction match. A scalar can never equal a vector.

Quantity	Type	Why
Distance	Scalar	Total path length — no direction
Displacement	Vector	Straight-line change in position, with direction
Speed	Scalar	Rate of distance — magnitude onlyQ NDA 2022 — speed is scalar, velocity is vector. The single most-tested line of this subtopic.
Velocity	Vector	Rate of displacement — has direction
Acceleration	Vector	Rate of change of velocity — has direction

Pair each scalar with its vector cousin: distance/displacement, speed/velocity. The vector member always carries a direction.

Practice this conceptself-check · 4 quick reps

Try it yourself

Classify each as scalar or vector: (i) the reading on a car's speedometer, (ii) a 3 km walk that ends back where it started, measured as displacement, (iii) the temperature of a room.

Practice — Level 1 (4 reps)

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

1.
Is speed a scalar or a vector?
2.
Is velocity a scalar or a vector?
3.
Is distance a scalar or a vector?
4.
Is displacement a scalar or a vector?

From the bank · past-year question

Example 1Kinematics and MotionEASY

Which one of the following statements about speed and velocity is correct?

[Q102 · Sep · 2022]

Speed is the scalar; velocity is the vector

The dominant distractor swaps them or calls both vectors. Speed is the magnitude of velocity (no direction); velocity carries a direction. A change in direction alone changes the velocity even when the speed is unchanged.

Concept 2 of 4

Distance vs displacement

Intuition

Distance is how far you actually travelled — the full length of the path, never negative, never decreasing. Displacement is the straight arrow from start to finish — it has a direction and can be zero even after a long trip if you return to the start.

Definition

Distance is the total path length covered (a scalar, always $\geq 0$ ). Displacement is the straight-line vector from the initial to the final position; its magnitude $\leq$ distance, with equality only for motion in a single straight line without reversal. For any closed loop (return to start), displacement $= 0$ while distance $=$ the path length.

Distance and displacement

|\vec{s}| \leq d \qquad \vec{s}_{\text{round trip}} = 0

ddistance (total path length, scalar)
\vec{s}displacement (start → finish, vector)

Worked example

A runner goes 300 m east, then 400 m north. Find the total distance and the magnitude of the displacement.

Distance = total path = $300 + 400 = 700$ m.
Displacement is the straight line from start to finish — the two legs are perpendicular, so use Pythagoras.
$|\vec{s}| = \sqrt{300^2 + 400^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500$ m.

Answer:Distance 700 m; displacement 500 m (north-east of the start).

Practice this conceptself-check · 4 quick reps

Try it yourself

A cyclist rides 2 km north then 2 km south back to the start. Find the distance and the displacement.

Practice — Level 1 (4 reps)

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

1.
A particle moves 6 m right then 8 m up. Magnitude of displacement?
2.
You walk once around a 400 m track and stop at the start. Displacement?
3.
Can displacement ever exceed distance?
4.
When are distance and displacement magnitude equal?

From the bank · past-year question

Example 2Kinematics and MotionEASY

A car starts from Bengaluru, goes 50 km in a straight line towards south, immediately turns around and returns to Bengaluru. The time taken for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip

[Q66 · Sep · 2019]

Round trip: distance is non-zero, displacement is zero

On any out-and-back trip the displacement is 0, so the average VELOCITY is 0 — but the distance and the average SPEED are not. The 50 km out-and-back PYQ tests exactly this: average velocity = 0, average speed = total distance / time.

Concept 3 of 4

Speed, velocity, and their averages

Intuition

Average speed divides total path by total time; average velocity divides the displacement vector by total time. On a round trip the average velocity is zero (no net displacement) while the average speed is not. Mixing the two is the classic trap.

Definition

Average speed $= \dfrac{\text{total distance}}{\text{total time}}$ (scalar). Average velocity $= \dfrac{\text{displacement}}{\text{total time}}$ (vector). Speed is the magnitude of instantaneous velocity. The two averages coincide only for straight-line motion without reversal.

Average speed and velocity

\bar{v}_{\text{speed}} = \dfrac{d}{t} \qquad \bar{v}_{\text{velocity}} = \dfrac{\vec{s}}{t}

dtotal distance
\vec{s}total displacement
ttotal time

Worked example

A car covers 120 km in 2 hours along a straight road in one direction. Find its average speed and the magnitude of its average velocity.

Straight-line motion, no reversal — distance = displacement magnitude = 120 km.
Average speed = $120 / 2 = 60$ km/h.
Average velocity magnitude = displacement / time = $120 / 2 = 60$ km/h.
They are equal here because the path is a single straight line.

Answer:Average speed 60 km/h; average velocity 60 km/h (equal, straight-line motion).

Practice this conceptself-check · 3 quick reps

Try it yourself

A car drives 50 km south, turns around, and returns to the start, the whole round trip taking 2 hours. Find the magnitude of its average velocity and its average speed.

Practice — Level 1 (3 reps)

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

1.
Odometer reads 2000 km then 2400 km after 8 h. Average speed?
2.
Round-trip average velocity is always what?
3.
Speed is the magnitude of what vector?

From the bank · past-year question

Example 3Kinematics and MotionEASY

Ram records the odometer readings of his car for the distance covered from 2000 km at the start of his journey and 2400 km at the end of the journey after 8 hours. What is the average speed of the car?

[Q83 · Sep · 2023]

Average speed is NOT |average velocity| in general

Average speed uses total distance; average velocity uses displacement. They match only for straight-line, no-reversal motion. Whenever a path curves or reverses, average speed > |average velocity|.

Concept 4 of 4

Position vector r(t) and net displacement

Intuition

The position vector r(t) lists a particle's x, y, z coordinates as functions of time, all in one expression. Differentiate it once to get velocity, again to get acceleration. When motion happens in two perpendicular legs, combine the legs with Pythagoras to get the net displacement.

Definition

The position vector $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}$ gives the location at time $t$ . Velocity is its time derivative, $\vec{v} = \dfrac{d\vec{r}}{dt}$ ; acceleration is $\vec{a} = \dfrac{d\vec{v}}{dt}$ . For two perpendicular displacement legs, the net displacement magnitude is $\sqrt{s_x^2 + s_y^2}$ .

Position vector and its derivatives

\vec{v} = \dfrac{d\vec{r}}{dt}, \quad \vec{a} = \dfrac{d\vec{v}}{dt}, \quad |\vec{s}_{\text{net}}| = \sqrt{s_x^2 + s_y^2}

\vec{r}position vector
\vec{v}velocity vector
\vec{a}acceleration vector

Worked example

A particle has

\vec{r}(t) = 3t^2\hat{i} + 4t\hat{j}

(metres, seconds). Find its velocity and acceleration at

t = 2

Differentiate once: $\vec{v} = \dfrac{d\vec{r}}{dt} = 6t\,\hat{i} + 4\,\hat{j}$ .
At $t = 2$ : $\vec{v} = 12\,\hat{i} + 4\,\hat{j}$ m/s.
Differentiate again: $\vec{a} = \dfrac{d\vec{v}}{dt} = 6\,\hat{i}$ m/s $^2$ (constant).

Answer:

\vec{v} = 12\,\hat{i} + 4\,\hat{j}

m/s;

\vec{a} = 6\,\hat{i}

m/s

^2

Practice this conceptself-check · 3 quick reps

Try it yourself

A vehicle starts from rest, accelerates at 2 m/s² east for 10 s, stops, then accelerates at

4\sqrt{2}

m/s² south for 10 s. Find the magnitude of its net displacement.

Practice — Level 1 (3 reps)

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

1.
Given $\vec{r} = 5t\,\hat{i}$ , what is the acceleration?
2.
Legs of 30 m east and 40 m north. Net displacement?
3.
Velocity is the derivative of which quantity?

From the bank · past-year question

Example 4Kinematics and MotionHARD

Vehicle from rest: accelerates at

2\,\text{m/s}^2

east for 10 s, stops, then accelerates at

4\sqrt{2}\,\text{m/s}^2

south for 10 s. Net displacement from starting point is

[Q98 · Sep · 2024]

Add perpendicular legs as vectors, not as numbers

A common error sums the two leg distances arithmetically (100 + 200√2 ≈ 383 m). Perpendicular displacements must be combined with Pythagoras: √(100² + (200√2)²) = 300 m. Only collinear legs add directly.

Force ∥ momentum needs both vectors checked

For

\vec{r} = \sqrt{3}t^2\hat{i} + \sqrt{2}t\hat{j} + \sqrt{5}\hat{k}

, the force is

m\vec{a} = 2\sqrt{3}m\,\hat{i}

(pure

\hat{i}

) while at

t=0

the momentum is

m\sqrt{2}\,\hat{j}

(pure

\hat{j}

). They are perpendicular at that instant — differentiate twice and compare directions, do not assume force is along motion.

Drill 1 more on position vector r(t) and net displacement

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 (3)

Distance vs displacement
Distance and displacement
$|\vec{s}| \leq d \qquad \vec{s}_{\text{round trip}} = 0$
Speed, velocity, and their averages
Average speed and velocity
$\bar{v}_{\text{speed}} = \dfrac{d}{t} \qquad \bar{v}_{\text{velocity}} = \dfrac{\vec{s}}{t}$
Position vector r(t) and net displacement
Position vector and its derivatives
$\vec{v} = \dfrac{d\vec{r}}{dt}, \quad \vec{a} = \dfrac{d\vec{v}}{dt}, \quad |\vec{s}_{\text{net}}| = \sqrt{s_x^2 + s_y^2}$

Reference tables (1)

Scalars vs vectors5 rows

Quantity	Type	Why
Distance	Scalar	Total path length — no direction
Displacement	Vector	Straight-line change in position, with direction
Speed	Scalar	Rate of distance — magnitude onlyQ NDA 2022 — speed is scalar, velocity is vector. The single most-tested line of this subtopic.
Velocity	Vector	Rate of displacement — has direction
Acceleration	Vector	Rate of change of velocity — has direction

Pair each scalar with its vector cousin: distance/displacement, speed/velocity. The vector member always carries a direction.

Watch out for (5)

Speed is the scalar; velocity is the vector
→ Scalars vs vectors
Round trip: distance is non-zero, displacement is zero
→ Distance vs displacement
Average speed is NOT |average velocity| in general
→ Speed, velocity, and their averages
Add perpendicular legs as vectors, not as numbers
→ Position vector r(t) and net displacement
Force ∥ momentum needs both vectors checked
→ Position vector r(t) and net displacement

Mastery check — 1 interleaved questions

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

Example 1Kinematics and MotionHARD

The position vector of a particle is given by

\vec{r} = \sqrt{3}t^2\hat{i} + \sqrt{2}t\hat{j} + \sqrt{5}\hat{k}

. Which one of the following statements is correct?

[Q57 · Apr · 2026]

Drill every past-year question on this subtopic

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

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