Motion in a Straight Line is the kinematics chapter that describes one-dimensional motion using position, distance, displacement, speed, velocity and acceleration, and the graphs and equations that connect them. It teaches the three equations of uniformly accelerated motion, free fall under gravity and relative velocity in one dimension. The chapter builds the foundation for all of mechanics and is a steady source of direct, solvable NEET questions.
Table of Contents
- Key Concepts — Position, displacement, velocity, speed, acceleration, graphs, relative velocity, free fall
- Weightage in Board & Entrance Exams
- Important Definitions
- Solved Examples
- Important Questions for Board Exams
- Quick Revision Points
Key Concepts
1. Position, Path Length and Displacement
To describe motion we first fix a reference point (origin) and a set of axes — together these form a frame of reference. The position of an object is its location measured from the origin along the chosen axis.
Path length (distance) is the total length of the actual path covered by the object. It is a scalar, is always positive, and can never be zero for a moving body.
Displacement (Δx) is the change in position: the straight-line distance from the start point to the end point, with direction. It is a vector and can be positive, negative, or zero.
Δx = x₂ − x₁
- If you walk 4 m east and 3 m back west, path length = 7 m but displacement = 1 m east.
- For a round trip, displacement is zero but path length is not.
- Magnitude of displacement ≤ path length, always.
2. Average Velocity and Average Speed
Average velocity is the displacement divided by the time interval. It is a vector, in the direction of displacement.
v̄ = Δx/Δt = (x₂ − x₁)/(t₂ − t₁)
Average speed is the total path length divided by the total time taken. It is a scalar and is always positive for a moving body.
Average speed = total path length / total time
- SI unit: m/s for both.
- Average speed ≥ magnitude of average velocity (they are equal only when motion is along a straight line in one direction).
3. Instantaneous Velocity and Speed
Instantaneous velocity is the velocity of a body at a particular instant of time. It is the limit of average velocity as the time interval approaches zero.
v = lim (Δt→0) Δx/Δt = dx/dt
- It is the slope of the tangent to the position-time graph at that instant.
- The magnitude of instantaneous velocity equals the instantaneous speed.
- The speedometer of a car shows instantaneous speed.
4. Acceleration
Acceleration is the rate of change of velocity with time. It is a vector and tells us how quickly velocity changes.
Average acceleration ā = Δv/Δt and instantaneous acceleration a = dv/dt = d²x/dt²
- SI unit: m/s².
- If velocity and acceleration point the same way, the body speeds up; if opposite, it slows down (retardation/deceleration).
- Acceleration is the slope of the velocity-time graph.
5. Kinematic Equations for Uniformly Accelerated Motion
When acceleration a is constant, the motion is called uniformly accelerated motion, and three standard equations connect initial velocity u, final velocity v, acceleration a, time t, and displacement s.
| Equation | What it connects |
|---|---|
| v = u + at | velocity and time (no s) |
| s = ut + ½at² | displacement and time (no v) |
| v² = u² + 2as | velocity and displacement (no t) |
A useful fourth result — the distance covered in the nth second:
sₙ = u + ½a(2n − 1)
Note: These equations are valid only when acceleration is constant. Take care with signs — choose one direction as positive and stick to it throughout the problem.
6. Position-Time Graphs
A position-time (x–t) graph plots position on the y-axis against time on the x-axis. Its slope at any point gives the instantaneous velocity.
[DIAGRAM: x–t graph — a horizontal line means the body is at rest; a straight slanted line means uniform velocity; an upward curve (increasing slope) means acceleration.]
- Straight horizontal line: object at rest (velocity = 0).
- Straight slanted line: uniform velocity (constant slope).
- Curved line: non-uniform velocity — accelerated motion.
- A steeper slope means a greater speed.
7. Velocity-Time Graphs
A velocity-time (v–t) graph plots velocity against time. It is one of the most powerful tools in kinematics because both its slope and its area carry meaning.
- Slope of the v–t graph = acceleration.
- Area under the v–t graph = displacement.
- A straight slanted line means uniform acceleration; a horizontal line means uniform velocity (zero acceleration).
[DIAGRAM: v–t graph for uniform acceleration — a straight line rising from u to v; the area under it (a trapezium) equals the displacement s = ut + ½at².]
The kinematic equations can actually be derived from the v–t graph: v = u + at comes from the slope, and s = ut + ½at² comes from the area.
8. Relative Velocity in One Dimension
The relative velocity of object A with respect to object B is the velocity of A as seen by an observer moving with B.
v(AB) = v(A) − v(B)
- Same direction: relative velocity = v(A) − v(B) (small if speeds are close — two trains moving alongside seem slow relative to each other).
- Opposite directions: relative velocity = v(A) + v(B) (they approach quickly).
- The time to meet = relative displacement / relative velocity.
9. Motion Under Gravity (Free Fall)
Near the Earth’s surface, every freely falling body has a constant downward acceleration called acceleration due to gravity (g ≈ 9.8 m/s²), independent of its mass.
The kinematic equations apply directly, replacing a with g and choosing a sign convention (usually downward positive for a dropped body):
- v = u + gt
- h = ut + ½gt²
- v² = u² + 2gh
For a body thrown vertically upward with speed u: it stops momentarily at the top (v = 0), so the maximum height is H = u²/2g, the time to reach the top is t = u/g, and the total time of flight is 2u/g.
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Weightage in Board & Entrance Exams
| Exam | Typical Weightage | Most-Tested Areas |
|---|---|---|
| CBSE Board (Class 11) | 6–8 marks | Distance vs displacement, kinematic equations, v–t graphs, free fall |
| JEE Main / Advanced | 1–2 questions | Graphs, relative velocity, motion under gravity numericals |
| NEET | 1–2 questions | Kinematic equations, average velocity vs speed, free fall |
[TABLE: Question-type split — VSA (1 mark): definitions, scalar vs vector; SA (2–3 marks): graph interpretation, kinematic numericals; LA (5 marks): derivation of equations from v–t graph, free-fall problems.]
Important Definitions
| Term | Definition |
|---|---|
| Position | Location of an object relative to a chosen origin along an axis |
| Path length (distance) | Total length of the actual path covered; a scalar, always positive |
| Displacement | Change in position; shortest directed distance: Δx = x₂ − x₁ (a vector) |
| Average velocity | Displacement per unit time: v̄ = Δx/Δt (a vector) |
| Average speed | Total path length per unit time (a scalar) |
| Instantaneous velocity | Velocity at an instant: v = dx/dt |
| Acceleration | Rate of change of velocity: a = dv/dt = d²x/dt² |
| Uniform motion | Equal displacements in equal time intervals (constant velocity) |
| Relative velocity | Velocity of one body as seen from another: v(AB) = v(A) − v(B) |
| Acceleration due to gravity | Constant downward acceleration of a free-falling body: g ≈ 9.8 m/s² |
Solved Examples
Example 1
A car travels 60 km east in 1 hour, then 40 km west in the next hour. Find its average speed and average velocity.
Answer: Path length = 60 + 40 = 100 km in 2 h, so average speed = 100/2 = 50 km/h. Displacement = 60 − 40 = 20 km east, so average velocity = 20/2 = 10 km/h east.
Example 2
A body starts from rest and accelerates uniformly at 2 m/s² for 5 s. Find its final velocity and the distance covered.
Answer: v = u + at = 0 + 2 × 5 = 10 m/s. s = ut + ½at² = 0 + ½ × 2 × 5² = 25 m.
Example 3
A car moving at 20 m/s is brought to rest in 50 m by applying brakes. Find the retardation.
Answer: Using v² = u² + 2as: 0 = 20² + 2a(50). a = −400/100 = −4 m/s² (retardation of 4 m/s²).
Example 4
A stone is dropped from a tower 80 m high. Find the time taken to reach the ground and its velocity on impact. (g = 10 m/s²)
Answer: h = ut + ½gt² → 80 = 0 + ½ × 10 × t², so t² = 16, t = 4 s. v = u + gt = 0 + 10 × 4 = 40 m/s.
Example 5
A ball is thrown vertically upward with a velocity of 20 m/s. Find the maximum height reached and the total time of flight. (g = 10 m/s²)
Answer: H = u²/2g = 20²/(2 × 10) = 400/20 = 20 m. Time of flight = 2u/g = (2 × 20)/10 = 4 s.
Example 6
Two trains A and B move on parallel tracks at 60 km/h and 40 km/h in the same direction. Find the velocity of A relative to B, and relative to B if B moves in the opposite direction.
Answer: Same direction: v(AB) = 60 − 40 = 20 km/h. Opposite directions: v(AB) = 60 + 40 = 100 km/h.
Important Questions for Board Exams
1-Mark Questions (VSA)
- Distinguish between distance and displacement in one line.
- Can displacement be zero while distance is not? Give an example.
- What does the slope of a position-time graph represent?
- What does the area under a velocity-time graph represent?
- Is acceleration due to gravity dependent on the mass of the falling body?
2–3-Mark Questions (SA)
- Define average velocity and instantaneous velocity. How are they related?
- A body covers equal distances in equal time intervals along a straight line. What can you say about its velocity and acceleration?
- Draw and explain the velocity-time graph for a uniformly accelerated body starting from rest.
- Two cars approach each other on a straight road. Explain how to find the time before they meet using relative velocity.
5-Mark Questions (LA)
- Derive the three kinematic equations of uniformly accelerated motion using a velocity-time graph.
- A ball is thrown vertically upward. Derive expressions for its maximum height, time of ascent, and total time of flight.
- Explain position-time and velocity-time graphs for uniform and non-uniform motion, with sketches and interpretation of slope and area.
Quick Revision Points
- Distance is a scalar (path length); displacement is a vector: Δx = x₂ − x₁
- |Displacement| ≤ distance; for a round trip displacement = 0
- Average velocity = Δx/Δt; average speed = path length/time
- Average speed ≥ |average velocity|
- Instantaneous velocity v = dx/dt = slope of x–t graph
- Acceleration a = dv/dt = slope of v–t graph
- Kinematic equations (constant a): v = u + at; s = ut + ½at²; v² = u² + 2as
- Distance in nth second: sₙ = u + ½a(2n − 1)
- Area under v–t graph = displacement
- Relative velocity (1-D): v(AB) = v(A) − v(B)
- Free fall: replace a with g (≈9.8 m/s²); max height H = u²/2g; time of flight = 2u/g
Next Chapter: Chapter 3 — Motion in a Plane
Class 11 Physics · Chapter 3 – swipe through all 9 cards to understand the whole chapter.
Distance vs Displacement
Distance is the actual path length; displacement is the straight-line change in position.
Equal only if motion is straight without reversing
- Distance: scalar, always ≥ 0, never decreases
- Displacement: vector, can be +, − or 0
- Return to start ⇒ displacement = 0, distance ≠ 0
Speed vs Velocity
Speed is built from distance; velocity is built from displacement.
Closed loop: avg speed > 0 but avg velocity = 0
- Speed: scalar (never negative)
- Velocity: vector (sign = direction)
- 1 km h⁻1 = 5/18 m s⁻1
Instantaneous Velocity
The velocity right now is the limit of Δx/Δt as the interval shrinks to a point.
Its magnitude = instantaneous speed
- Speedometer reading
- Slope of position-time graph at a point
- Equal-time avg = (v1+v2)/2; equal-dist avg = 2v1v2/(v1+v2)
Acceleration
Acceleration is the rate of change of velocity, not just speeding up.
Unit: m s⁻2 ; a can be ≠ 0 even when v = 0
- a, v same sign ⇒ speeding up
- a, v opposite sign ⇒ slowing down
- Negative a ≠ slowing down (just points along −x)
Three Equations of Motion
For constant acceleration a, these link u, v, a, s and t.
Valid ONLY for constant a; pick +ve direction first
- s = displacement, all quantities signed
- Use the equation with your 3 knowns + 1 unknown
- Distance in nth second: sₙ = u + (a/2)(2n − 1)
Free Fall (Motion Under Gravity)
Free fall is the three equations with a = g ≈ 9.8 m s⁻2 (use 10 if told).
Thrown up: a = −g throughout, even at the top where v = 0
- Down taken +ve when object only falls
- Up-throw: u > 0 but a = −g
- Stopping distance = u2/2a (scales with speed2)
Position-Time (x-t) Graph
On an x-t graph the slope tells you the velocity.
From rest under uniform a: x ∝ t2 (parabola)
- Straight sloping line ⇒ uniform velocity
- Horizontal line ⇒ body at rest
- Negative slope ⇒ motion in −x direction
Velocity-Time (v-t) Graph
On a v-t graph slope = acceleration and area = displacement.
Area below the time axis counts as negative
- Slope of v-t = acceleration
- Straight line ⇒ uniform acceleration
- v-t can’t be vertical (infinite a is unphysical)
Relative Velocity (1-D)
The velocity of A as seen by B is found by subtracting their signed velocities.
Always assign signs before subtracting
- Same direction ⇒ speeds subtract
- Opposite directions ⇒ speeds add
- Time to meet = separation / |v_rel|
📝 Practice Motion in a Straight Line — 10 NEET PYQs
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- Motion in a Plane Class 11 Notes
- Laws of Motion Class 11 Notes
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Frequently Asked Questions
Distance is the total path length actually covered and is a scalar that is always positive. Displacement is the straight-line change in position from start to finish and is a vector that can be positive, negative or zero, so an object that returns to its start has zero displacement but non-zero distance.
For constant acceleration they are v = u + at, s = ut + half a t squared, and v squared = u squared + 2as, where u is initial velocity, v is final velocity, a is acceleration, s is displacement and t is time. They are valid only when acceleration is constant, and all the quantities must carry signs based on a chosen positive direction.
Speed is built from distance and is a scalar that is never negative, while velocity is built from displacement and is a vector whose sign shows direction. For a closed loop the average speed is positive but the average velocity is zero because the displacement is zero.
Yes. A ball thrown straight up has zero velocity for an instant at its highest point, but its acceleration is still g pointing downward the whole time. Acceleration is the rate of change of velocity, so it can be non-zero even when the velocity is momentarily zero.
Yes. It is part of the Class 11 Physics NEET syllabus and usually contributes about one to two direct questions, often on the equations of motion, free fall, graphs or relative velocity. Because the questions are formula-based and quick to solve, it is a high-yield scoring chapter.