Physics
Kinematics
Motion in a Straight Line

Kinematics · JEE & NEET Physics

Motion in a Straight Line: notes and previous year questions

Distance and displacement, speed and velocity, acceleration and its sign, and how calculus links position, velocity and acceleration.

56 JEE Main questions (2003–2026)
26 NEET questions (2000–2026)

Motion in a Straight Line in short

In one dimension, choose an origin and a positive direction; the sign of x, v and a then gives the direction.
Distance is the whole path (a scalar, never negative); displacement is the change in position (a vector).
Distance ≥ |displacement|, equal only when the motion never turns back; a round trip has zero displacement.
Average speed uses total distance; average velocity uses displacement; average speed ≥ |average velocity|.

1Distance and displacement

A batter taps the ball and runs two quick runs. The pitch is about 20 m between the creases, so they run 20 m + 20 m = 40 m. Yet after two runs they stand exactly where they started. The distance (the whole path) is 40 m; the displacement (the change in position) is 0 m.

Motion along a single line is called rectilinear or one-dimensional motion. Choose an origin and a positive direction (say, to the right is $+x$ ). Then position is one number $x$ , which can be positive or negative, and velocity and acceleration carry a $+$ or $-$ sign. In one dimension, the sign is the direction.

	Distance	Displacement
What it is	the total path length	the change in position
Type	scalar	vector
Sign	never negative	$+$ , $-$ or $0$
Formula	add up every part of the path	$\vec{s} = \vec{r}_f - \vec{r}_i$

\text{distance} \ge |\text{displacement}|

Equal only when the motion never turns back.

2Speed and velocity

Speed is how fast distance is covered; velocity is how fast displacement changes. Both are measured in m/s. Speed is never negative; velocity can be positive, negative or zero.

\text{average speed} = \frac{\text{total distance}}{\text{total time}}

\vec{v}_{\text{avg}} = \frac{\Delta\vec{s}}{\Delta t}

The instantaneous velocity is the average velocity over a shorter and shorter time. On a position–time graph, the chord from P to Q turns into the tangent at P:

v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}

For x = t², the slope at t = 2 s is 4 m/s.

Situation	Speed and velocity
Uniform motion in one direction	speed = \|velocity\|
Motion with a reversal	average speed > \|average velocity\|
A complete round trip	average speed > 0, average velocity = 0

3Average speed shortcuts

Equal distances at two speeds (the first half of the distance at $v_1$ , the second half at $v_2$ ). Each half of length $d$ takes $d/v_1$ and $d/v_2$ :

v_{\text{avg}} = \frac{2d}{d/v_1 + d/v_2} = \frac{2v_1v_2}{v_1 + v_2}

The harmonic mean.

Equal times at two speeds (half the time at $v_1$ , half at $v_2$ ):

v_{\text{avg}} = \frac{v_1 + v_2}{2}

The arithmetic mean.

For unequal speeds the harmonic mean is always less than the arithmetic mean: the slower part lasts longer and pulls the average down. Memory aid: same distance → harmonic, same time → arithmetic.

4Acceleration and its sign

Acceleration is the rate of change of velocity, measured in m/s²:

\vec{a} = \frac{d\vec{v}}{dt}

Positive acceleration: velocity increasing in the positive direction.
Negative acceleration: velocity decreasing (often called deceleration or retardation when the object slows down).
Zero acceleration: constant velocity, uniform motion.

Directions of v and a	Motion
Same direction	speeding up
Opposite directions	slowing down

5The calculus ladder

Position, velocity and acceleration are rungs of one ladder. Going down ( $x \to v \to a$ ) you differentiate (take slopes). Going up you integrate (take areas).

Direction	Operation	Relation
$x \to v$	differentiate	$v = dx/dt$
$v \to a$	differentiate	$a = dv/dt$
$a \to v$	integrate	$v = u + \int_0^t a\,dt$
$v \to x$	integrate	$x = x_0 + \int_0^t v\,dt$

By the chain rule there is a third form of acceleration, without time. Use it whenever velocity is given as a function of position:

a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\,\frac{dv}{dx}

6Turning points and distance

When a particle reverses, distance and displacement differ, and the distance is not $|x_{\text{final}} - x_{\text{initial}}|$ . The method:

Solve $v = 0$ to find the turning points.
Find the position at the start, at each turning point and at the end.
Add the size of every leg.

Summary

Key ideas

In one dimension, choose an origin and a positive direction; the sign of x, v and a then gives the direction.
Distance is the whole path (a scalar, never negative); displacement is the change in position (a vector).
Distance ≥ |displacement|, equal only when the motion never turns back; a round trip has zero displacement.
Average speed uses total distance; average velocity uses displacement; average speed ≥ |average velocity|.
Instantaneous velocity is the slope of the position–time graph: v = dx/dt; its size is the instantaneous speed.
Equal distances at two speeds give the harmonic mean; equal times give the arithmetic mean.
The harmonic mean is always smaller, so the simple average of two speeds is a trap for equal distances.
Acceleration is dv/dt; an object speeds up when v and a point the same way and slows down when they are opposite.
Differentiate down the ladder x → v → a; integrate up; use a = v dv/dx when v is given in terms of x.
The constant-acceleration equations fail when a changes; integrate instead.
For the distance with reversals, split the motion at the points where v = 0.

Every equation

Displacement: $\vec{s} = \vec{r}_f - \vec{r}_i$
Average speed: $\text{average speed} = \frac{\text{total distance}}{\text{total time}}$
Average velocity: $\vec{v}_{\text{avg}} = \frac{\Delta\vec{s}}{\Delta t}$
Instantaneous velocity: $v = \frac{dx}{dt}$
Equal distances: $v_{\text{avg}} = \frac{2v_1v_2}{v_1 + v_2}$
Equal times: $v_{\text{avg}} = \frac{v_1 + v_2}{2}$
Acceleration: $a = \frac{dv}{dt}$
Acceleration from v(x): $a = v\,\frac{dv}{dx}$
Velocity from acceleration: $v = u + \int_0^t a\,dt$
Position from velocity: $x = x_0 + \int_0^t v\,dt$

Previous year questions with solutions

Real JEE and NEET questions on motion in a straight line. Try each one before you open the solution.

Q1JEE Main 2026Numerical answer

From $18\ \mathrm{m}$ height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity is equal to the magnitude of acceleration due to gravity (in the same set of units) is __________ $\mathrm{m}$ . (Take $g=10\ \mathrm{m/s^2}$ and neglect the air resistance.)

Show answer and solution

Answer: 13

The question asks where the speed reaches the number $10$ , since $g=10$ in these units. From $v^{2}=u^{2}+2gy$ with $u=0$ : $100=2(10)y$ , so $y=5\ \mathrm{m}$ of falling. The ball started at $18\ \mathrm{m}$ , so it is then $18-5=13\ \mathrm{m}$ above the ground.

Q2NEET 2026One correct option

Consider a particle moving along a straight line, whose position as a function of time is given by $s(t) = \alpha t^{2} - \beta t + \gamma$ , where $\alpha = 1\ \mathrm{m\,s^{-2}}$ , $\beta = 6\ \mathrm{m\,s^{-1}}$ and $\gamma = 5\ \mathrm{m}$ . The average speed of the particle, in $\mathrm{m\,s^{-1}}$ , from $t=0$ to $t=6\ \mathrm{s}$ is:

A $0$
B $12$
C $6$
D $3$

Show answer and solution

Answer: Option D

$v = \dfrac{ds}{dt} = 2t-6$ , which is zero at $t=3\ \mathrm{s}$ — the particle turns round in the middle of the interval. Distance, not displacement, is wanted, so take the two halves separately: each covers $\tfrac{1}{2}(3)(6) = 9\ \mathrm{m}$ , giving $18\ \mathrm{m}$ in $6\ \mathrm{s}$ , so $3\ \mathrm{m\,s^{-1}}$ . Average $\textit{velocity}$ here would have been zero.

Q3JEE Main 2026One correct option

Water drops fall from a tap on the floor, $5\ \mathrm{m}$ below, at regular intervals of time, the first drop striking the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground, at the instant when the first drop strikes the ground, is: ( $g=10\ \mathrm{m/s^2}$ )

A $3.8\ \mathrm{m}$
B $4.0\ \mathrm{m}$
C $4.2\ \mathrm{m}$
D $2.5\ \mathrm{m}$

Show answer and solution

Answer: Option C

Six drops means five equal intervals $T$ , and the first drop's whole fall takes $5T$ . From $5=\tfrac{1}{2}(10)(5T)^{2}$ comes $T=0.2\ \mathrm{s}$ . The fourth drop started three intervals later, so at that instant it has been falling for $5T-3T=0.4\ \mathrm{s}$ and has dropped $\tfrac{1}{2}(10)(0.4)^{2}=0.8\ \mathrm{m}$ , leaving it $5-0.8=4.2\ \mathrm{m}$ up.

Practice questions, easy to hard

Three questions from the motion in a straight line practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Average speed is the total distance divided by the total time — one number for a whole journey, however the journey actually went.

A cyclist covers $900\ \mathrm{m}$ in $300\ \mathrm{s}$ . What is her average speed, in $\mathrm{m/s}$ ?

Show answer and solution

Answer: 3

$900 / 300 = 3\ \mathrm{m/s}$ . It does not claim she rode at $3\ \mathrm{m/s}$ at any moment — it is the steady speed that would have taken exactly as long.

Q5Numerical answer

The rise takes $u/g$ , and the fall back to the throwing point takes exactly as long again.

A ball thrown up at $20\ \mathrm{m/s}$ returns to the hand after how many seconds? Take $g=10\ \mathrm{m/s^2}$ .

Show answer and solution

Answer: 4

Time up is $u/g=20/10=2\ \mathrm{s}$ , and the journey down mirrors it, so the whole flight is $2u/g=4\ \mathrm{s}$ . The two halves are mirror images: same distance, same acceleration, opposite direction.

Q6Numerical answer

Going the other way — from velocity back to position — undoes the differentiation, and is the area under the $v$ – $t$ curve you already know. For $v = 3t^{2}$ the position that differentiates to it is $x = t^{3}$ .

A particle starts at the origin with $v = 3t^{2}$ . Where is it at $t = 2\ \mathrm{s}$ , in $\mathrm{m}$ ?

Show answer and solution

Answer: 8

Undoing $v = 3t^{2}$ gives $x = t^{3}$ plus a constant, and starting at the origin makes that constant zero. So $x(2) = 8\ \mathrm{m}$ . Check it forwards: $\dfrac{d}{dt}t^{3} = 3t^{2}$ .