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

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

The same idea in a real exam question:

NDA · Sep 2022 · Q102Easy

Example 1 · Kinematics and Motion · Vectors and Position

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

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

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.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q66Easy

Example 2 · Kinematics and Motion · Vectors and Position

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

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

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.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q83Easy

Example 3 · Kinematics and Motion · Vectors and Position

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?

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

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

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2024 · Q98Hard

Example 4 · Kinematics and Motion · Vectors and Position

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

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

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