When can I use v equals u plus at and the other equations of motion?

Only when the acceleration is constant throughout the motion. For variable acceleration, integrate: v equals integral of a dt and x equals integral of v dt.

What does the slope of a position-time graph represent?

The slope of an x-t graph equals the instantaneous velocity: v equals dx/dt. Horizontal segment means at rest; positive slope means motion in the plus x direction; negative slope means motion in the minus x direction.

Motion In One Dimension

Q: What is the difference between distance and displacement in motion in one dimension?

Distance is the total path length travelled (scalar, always positive). Displacement is the change in position from initial to final point in a specified direction (vector, can be positive, negative or zero). In straight-line motion without reversal they are equal in magnitude; otherwise displacement is smaller.

Q: Why is uniform circular motion called uniformly accelerated even though speed is constant?

Velocity is a vector. Its direction changes continuously in circular motion even when its magnitude (speed) stays constant. A changing velocity means non-zero centripetal acceleration v squared over r directed toward the centre.

Q: Can average velocity be zero while average speed is non-zero?

Yes. For a body that returns to its starting point, displacement is zero so average velocity is zero, but total distance is non-zero so average speed is non-zero.

Q: How do I read the area under a v-t graph physically?

The area between the velocity curve and the time axis equals the displacement during that interval. Areas above the axis are positive displacement; areas below are negative. Total distance is the sum of absolute values.

Q: What is relative velocity and why does it matter?

Relative velocity of A with respect to B is V_A minus V_B. It matters because problems about two moving objects depend on how fast one closes on the other, not on absolute velocities.

Q: On a smooth incline, does the final velocity at the bottom depend on the angle?

No - it depends only on the vertical height h: v equals square root of 2gh. The angle affects the time taken and the acceleration along the plane but not the final speed.

PhysicsKinematicsFor JEE aspirants

Motion in one dimension is the motion of an object along a straight line. Its position is described by a single coordinate (usually x), and vector quantities like displacement, velocity, and acceleration have only one non-zero component. This concept covers the core kinematic ideas used across JEE and NEET Physics: distance versus displacement, average and instantaneous speed and velocity, average and instantaneous acceleration, the three equations of uniformly accelerated motion, motion on an inclined plane, x-t, v-t, and a-t graphs, and relative velocity in a straight line.

Key Formulas - Quick Reference

Average speed: $v_{a v} = \frac{Δ s}{Δ t}$ Average velocity: $v_{a v} = \frac{Δ r}{Δ t}$
Instantaneous velocity: $v = \frac{d r}{d t}$ Instantaneous acceleration: $a = \frac{d v}{d t}$
Equations of motion (constant $a$ ): $v = u + a t$
$s = u t + \frac{1}{2} a t^{2}$
$v^{2} = u^{2} + 2 a s$
Distance in $n$ -th second: $s_{n} = u + \frac{a}{2} (2 n - 1)$
On incline (from rest): $v = (g sin θ) t$ , $s = \frac{1}{2} (g sin θ) t^{2}$ , $v = 2 g h$
Displacement = area under v-t graph; change in velocity = area under a-t graph
Relative velocity: $V_{A B} = V_{A} - V_{B}$

1. Kinematics vs Dynamics

Kinematics is the branch of mechanics that describes the motion of an object without asking why it moves. Dynamics studies the motion together with its cause (the forces producing it). This concept deals with kinematics only.

1.1 Motion in one dimension

Motion of an object along a straight line is called one-dimensional (1-D) motion. The position of the particle is described by a single variable (say $x$ ). All vector quantities like position, velocity, displacement, and acceleration have only one non-zero component.

1.2 Motion in two and three dimensions (for contrast)

Motion in a plane is called two-dimensional (2-D) motion - vectors have two non-zero components. Motion in space is called three-dimensional (3-D) motion - vectors have three non-zero components. These are treated in a separate concept.

2. Distance and Displacement

Displacement is the change in position of a body in a specified direction. It is the shortest straight-line vector from the initial to the final position. It is a vector quantity with SI unit metre (m).

Distance is the total length of the actual path traversed by the body. It is a scalar quantity with SI unit metre (m).

Displacement can be positive, negative, or zero; distance is always positive.
If a particle moves in a straight line without changing direction, magnitude of displacement equals the distance.
Otherwise: $∣ displacement ∣ < distance$ .

Solved Example 1

A particle moves along a circle of radius

R

from A to B (half cycle) and then back from B to A. Find the distance and displacement for the half cycle and for the full cycle.

Figure 1: Distance versus displacement for motion on a circular path (semicircle A→B and full circle A→A).

Solution:

Quantity	Half cycle (A→B or B→A)	Full cycle (A→A via B)
Distance	$π R$	$2 π R$
Displacement	$2 R$	$0$
Direction of displacement	A→B or B→A (as case may be)	Zero vector

3. Average Speed and Average Velocity

The average speed is the ratio of total distance travelled to the total time taken. The average velocity is the ratio of net displacement to the total time taken.

Figure 2: For a curved path, distance is the path length ABC while displacement is the straight-line vector from A to C.

If a particle moves from A to C along a curved path ABC, the distance $Δ s$ is the path length ABC, while the displacement is:

Δ r = r_{C} - r_{A}

Thus in time $Δ t$ :

v_{a v} = \frac{Δ s}{Δ t}, v_{a v} = \frac{Δ r}{Δ t}

4. Instantaneous Speed and Velocity

Instantaneous speed and velocity are defined at a particular instant as limits:

v = Δ t \to 0 lim \frac{Δ s}{Δ t} = \frac{d s}{d t}, v = Δ t \to 0 lim \frac{Δ r}{Δ t} = \frac{d r}{d t}

5. Average and Instantaneous Acceleration

Average acceleration is the change in velocity over a time interval:

a_{a v} = \frac{Δ v}{Δ t}

Instantaneous acceleration is the rate of change of velocity at that instant:

a = Δ t \to 0 lim \frac{Δ v}{Δ t} = \frac{d v}{d t}

Solved Example 2

A particle moves along a semicircular path from A to B (diameter

2 R

) in time

T

.
(a) Determine the average speed of the particle.
(b) Determine the average velocity of the particle.

Figure 3: Semicircular path from A to B - distance πR, displacement 2R.

Solution:

(a) Distance = arc length = $π R$ , so average speed $= \frac{π R}{T}$ .

(b) Displacement = straight line A to B = $2 R$ , so magnitude of average velocity $= \frac{2 R}{T}$ .

6. Equations of Motion (Constant Acceleration)

For an object moving in a straight line with constant acceleration $a$ :

(a)

v = u + a t

(b)

s = u t + \frac{1}{2} a t^{2}

(c)

v^{2} = u^{2} + 2 a s

Here $u$ = initial velocity (take $u = 0$ if the body starts from rest), $v$ = final velocity, $s$ = displacement in time $t$ , and $a$ = acceleration. Use a + sign for acceleration and a − sign for retardation (deceleration).

The displacement of the body in the $n$ -th second is given by:

s_{n} = u + \frac{a}{2} (2 n - 1)

6.1 Motion on an inclined plane

Figure 4: Body sliding down a frictionless incline - acceleration along plane = g sinθ.

A body of mass $m$ slides down a frictionless plane inclined at angle $θ$ to the horizontal. At $t = 0$ , the body is at rest at the top, so $u = 0$ and the acceleration along the plane is $a = g sin θ$ .

(a) $v = (g sin θ) t$
(b) $s = \frac{1}{2} (g sin θ) t^{2}$
(c) $v^{2} = 2 (g sin θ) s$

If the length of the plane is $s$ and vertical height is $h$ , then $sin θ = h / s$ , so:

t = \frac{1}{sin θ} \frac{2 h}{g}, v = 2 g h

Note: The velocity at the bottom depends only on the height

h

, not on the angle. The time taken, however, depends on the angle - a steeper incline gets the body down faster for the same height.

Solved Example 3

A ball is projected vertically upward with a velocity of 20 m/s. Find the distance travelled in the first three seconds. (Take

g = 10 m/s^{2}

Solution:

The direction of velocity reverses at the highest point. Using $v = u + a t$ with $v = 0$ :

0 = 20 - 10 t \Rightarrow t = 2 s

Distance in first 2 s (upward), where distance equals displacement because velocity does not change sign:

S_{1} = 20 \times 2 - \frac{1}{2} \times 10 \times 4 = 40 - 20 = 20 m

Distance in the next 1 s (downward from rest at the top):

S_{2} = \frac{1}{2} \times 10 \times 1^{2} = 5 m

Total distance in first 3 s = $20 + 5 = 25$ m.

Solved Example 4

A particle moves along the x-axis according to

x = 4 t - t^{2}

. Find the distance travelled from

t = 0

s to

t = 3

Solution:

Velocity: $v = d x / d t = 4 - 2 t$ . It vanishes at $t = 2$ s and changes sign. Split the integral at $t = 2$ :

Distance = \int_{0}^{2} (4 - 2 t) d t + \int_{2}^{3} (4 - 2 t) d t

= ∣ [4 t - t^{2}]_{0}^{2} ∣ + ∣ [4 t - t^{2}]_{2}^{3} ∣ = ∣4∣ + ∣ - 1∣ = 4 + 1 = 5 m

Solved Example 5

A body has uniform acceleration of

10 m/s^{2}

and a velocity of

100 m/s

. In what time will its velocity be doubled?

Solution:

Use $v = u + a t$ : $200 = 100 + 10 t \Rightarrow t = 10$ s.

Solved Example 6

A particle moves with uniform acceleration from A to B along a straight line with velocities

v_{1}

at A and

v_{2}

at B. If C is the midpoint of AB, find the velocity

v

at C.

Solution:

Let AB $= x$ . From $v^{2} = u^{2} + 2 a s$ applied over A to B:

v_{2}^{2} = v_{1}^{2} + 2 a x \Rightarrow 2 a x = v_{2}^{2} - v_{1}^{2}

Now apply the same equation over A to C (distance $x /2$ ):

v^{2} = v_{1}^{2} + 2 a (\frac{x}{2}) = v_{1}^{2} + a x = v_{1}^{2} + \frac{v _{2}^{2} - v _{1}^{2}}{2} = \frac{v _{1}^{2} + v _{2}^{2}}{2}

v = \frac{v _{1}^{2} + v _{2}^{2}}{2}

Solved Example 7

A body travels with a uniform acceleration of

2 m/s^{2}

and initial velocity

10 m/s

. What is its velocity after 4 s?

Solution:

$v = u + a t = 10 + 2 \times 4 = 18 m/s$ .

7. x-t, v-t and a-t Graphs

7.1 How x, v, and a vary with time

Type of motion	Displacement (x)	Velocity (v)	Acceleration (a)
At rest	Constant (horizontal line)	Zero	Zero
Constant velocity	Straight line, non-zero slope	Constant (horizontal line)	Zero
Constant acceleration	Parabola opening upward	Straight line, positive slope	Constant, positive
Constant deceleration	Parabola opening downward	Straight line, negative slope	Constant, negative

7.2 Displacement from a v-t graph

Figure 5: The area under a v-t graph equals the displacement.

The displacement during the interval $[t_{i}, t_{f}]$ equals the area bounded by the velocity curve and the time axis.

For a series of constant-velocity segments: $Δ x = \sum v_{i} Δ t_{i}$ .
For a smoothly varying curve: $Δ x = \int_{t_{i}}^{t_{f}} v (t) d t$ .

7.3 Change in velocity from an a-t graph

Similarly, given an acceleration versus time graph, the change in velocity between $t = t_{i}$ and $t = t_{f}$ equals the area bounded by the acceleration curve and the time axis.

Constant $a$ : $Δ v = a_{0} Δ t$ .
Varying $a$ : $Δ v = \int_{t_{i}}^{t_{f}} a (t) d t$ .

Solved Example 8

The velocity-time graph of a particle moving along a straight line goes from

v = 0

t = 0

v = 4

m/s at

t = 2

s (linear), then decreases linearly from

v = 4

m/s at

t = 2

s to

v = 2

m/s at

t = 3

s. The particle starts from

x = - 4

m.
(i) Sketch the a-t and x-t graphs.
(ii) Find the displacement at

t = 3

Solution:

(i) During $0 < t < 2$ s, $a = 4/2 = 2 m/s^{2}$ (constant). During $2 < t < 3$ s, $a = (2 - 4) /1 = - 2 m/s^{2}$ (constant, deceleration). The x-t graph is a parabola opening up in the first phase and a parabola opening down in the second phase.

(ii) Position at $t = 3$ s (starting from $x_{0} = - 4$ m):

x = - 4 + area under 0-2s \frac{1}{2} (2) (4) + area under 2-3s (\frac{4 + 2}{2}) (3 - 2) = - 4 + 4 + 3 = 3 m

8. Relative Velocity in a Straight Line

When two objects move along the same straight line, we compare their motion in terms of their relative velocity. If A and B move with velocities $V_{A}$ and $V_{B}$ respectively, then the relative velocity of A with respect to B is:

V_{A B} = V_{A} - V_{B}

Similarly, the relative velocity of B with respect to A is:

V_{B A} = V_{B} - V_{A}

Key points

Relative velocity of a particle = velocity of the particle − velocity of the reference object.
Same direction: relative velocity $= V_{A} - V_{B}$ (subtract magnitudes).
Opposite directions: relative velocity $= V_{A} + V_{B}$ (add magnitudes).
Two objects moving with the same velocity appear stationary relative to each other.

Solved Example 9

The position of a particle moving on a straight line is given by

x = 12 + 18 t + 9 t^{2}

metre. Find its velocity at

t = 2

Solution:

$v = d x / d t = 18 + 18 t$ . At $t = 2$ s: $v = 18 + 18 \times 2 = 54 m/s$ .

Common Mistakes to Avoid

Watch out

Confusing distance and displacement. A car returning to its starting point has zero displacement but non-zero distance. Displacement is a vector; distance is a scalar.
Forgetting to split the integral when velocity changes sign. For distance travelled (not displacement), split the integration at the instant $v = 0$ and add the absolute values of the segments.
Using the sign of $a$ carelessly. Acceleration is a vector - take the same sign convention as displacement/velocity throughout the problem, or you will get wrong results.
Confusing velocity at the highest point with acceleration. At the top of a vertical throw, velocity is zero for an instant but acceleration is still $g$ downward.
Applying $v = 2 g h$ on a rough incline. The result comes from energy conservation without friction. If friction is present, kinetic energy is dissipated - use a full force analysis.
Assuming average velocity equals arithmetic mean of $u$ and $v$ . This holds only for uniform acceleration; for other cases use $v_{a v} = Δ r /Δ t$ .

Frequently Asked Questions

Q1. What is the difference between distance and displacement in motion in one dimension?

Distance is the total length of the actual path travelled and is a scalar (always positive). Displacement is the change in position from the initial to the final point in a specified direction and is a vector (can be positive, negative, or zero). In straight-line motion without reversal, they are equal in magnitude; otherwise displacement is smaller than distance.

Q2. Why is uniform circular motion called "uniformly accelerated" even though the speed is constant?

Because velocity is a vector, its direction changes continuously in circular motion even when its magnitude (speed) is constant. A change in velocity means non-zero acceleration - specifically the centripetal acceleration $v^{2} / r$ directed toward the centre. This is discussed further in the Circular Motion concept.

Q3. Can average velocity be zero while average speed is non-zero?

Yes. For a body that returns to its starting point (e.g., a complete circle), displacement is zero, so average velocity is zero. But the total distance travelled is non-zero, so average speed is non-zero. This scenario is a favourite in JEE and NEET conceptual questions.

Q4. When can I use the equations $v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ , and $v^{2} = u^{2} + 2 a s$ ?

Only when the acceleration is constant throughout the motion (uniformly accelerated motion). For variable acceleration, you must integrate: $v = \int a d t$ and $x = \int v d t$ . Free-fall under gravity (no air resistance) and motion on a smooth incline are two standard settings where the equations apply.

Q5. How do I read the area under a v-t graph physically?

The area between the velocity curve and the time axis equals the displacement during that interval. Areas above the time axis count as positive displacement; areas below count as negative. If you want the total distance (a scalar), take the absolute values of each region separately and sum them.

Q6. What is relative velocity and why does it matter?

Relative velocity is the velocity of one object measured from another moving reference frame: $V_{A B} = V_{A} - V_{B}$ . It matters because problems about two cars, two trains, or a boat in a river depend on how fast one object closes on or moves away from the other - not on their absolute velocities.

Q7. On a smooth incline, does the final velocity at the bottom depend on the angle?

No - the final velocity depends only on the vertical height $h$ : $v = 2 g h$ . The angle affects the time taken (steeper = faster) and the acceleration along the plane ( $g sin θ$ ), but not the final speed. This follows directly from energy conservation on a frictionless surface.

Q8. What does the slope of a position-time (x-t) graph represent?

The slope of the x-t graph at any point equals the instantaneous velocity at that instant: $v = d x / d t$ . A horizontal segment means the particle is at rest; a positive slope means motion in the +x direction; a negative slope means motion in the -x direction. Similarly, the slope of a v-t graph is the acceleration.

Previous year questions on Motion In One Dimension

23 questions from past papers, each with a step-by-step solution.

Show all 23 questions

Ready to master Kinematics?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. Kinematics vs Dynamics

1.1 Motion in one dimension

1.2 Motion in two and three dimensions (for contrast)

2. Distance and Displacement

3. Average Speed and Average Velocity

4. Instantaneous Speed and Velocity

5. Average and Instantaneous Acceleration

6. Equations of Motion (Constant Acceleration)

6.1 Motion on an inclined plane

7. x-t, v-t and a-t Graphs

7.1 How x, v, and a vary with time

7.2 Displacement from a v-t graph

7.3 Change in velocity from an a-t graph

8. Relative Velocity in a Straight Line

Key points

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between distance and displacement in motion in one dimension?

Q2. Why is uniform circular motion called "uniformly accelerated" even though the speed is constant?

Q3. Can average velocity be zero while average speed is non-zero?

Q4. When can I use the equations v=u+at, s=ut+21​at2, and v2=u2+2as?

Q5. How do I read the area under a v-t graph physically?

Q6. What is relative velocity and why does it matter?

Q7. On a smooth incline, does the final velocity at the bottom depend on the angle?

Q8. What does the slope of a position-time (x-t) graph represent?

Previous year questions on Motion In One Dimension

Ready to master Kinematics?

Q4. When can I use the equations $v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ , and $v^{2} = u^{2} + 2 a s$ ?