Chapter Introduction
Motion in a Straight Line is the first kinematics chapter and one of the most frequently tested chapters in CBSE board exams (8–10 marks). It introduces the language of motion — displacement, velocity, acceleration — and the three kinematic equations that you will use throughout Class 11 and 12 Physics. Graph interpretation is a favourite exam topic. Master this chapter and every subsequent mechanics chapter becomes significantly easier.
What this guide covers: frame of reference, displacement vs. distance, speed and velocity (average and instantaneous), acceleration, three kinematic equations, graph interpretation (x-t, v-t, a-t), relative velocity, and free fall under gravity — all with worked numericals.
Key Definitions
- Frame of reference: A coordinate system with a clock to measure position and time. All motion is described relative to a chosen frame.
- Rest and motion: An object is at rest if its position does not change with time relative to a reference point; it is in motion if its position changes.
- Position: Location of a particle along a chosen axis, measured from a reference point (origin). It is a scalar in 1D but has a sign (+/−) indicating direction.
- Path length (distance): Total length of the actual path traversed by the particle. Always positive, scalar.
- Displacement: Change in position = final position − initial position. Vector quantity; can be positive, negative, or zero.
- Speed: Rate of change of distance with time. Scalar, always positive. Average speed = total distance / total time.
- Velocity: Rate of change of displacement with time. Vector; can be positive or negative. Average velocity = displacement / total time.
- Instantaneous velocity: Velocity at a particular instant. v = lim<sub>Δt→0</sub> (Δx/Δt) = dx/dt. Equal to the slope of the position-time graph at that point.
- Instantaneous speed: Magnitude of instantaneous velocity. Always positive.
- Acceleration: Rate of change of velocity with time. a = dv/dt. Vector; positive when speeding up in the positive direction, negative when slowing down (retardation/deceleration).
- Uniform acceleration: Acceleration is constant in magnitude and direction throughout the motion.
- Relative velocity: Velocity of one object as observed from another moving object. v<sub>AB</sub> = v<sub>A</sub> − v<sub>B</sub>.
- Free fall: Motion under gravity alone (neglecting air resistance). Acceleration = g ≈ 9.8 m s<sup>-2</sup> (downward).
Detailed Concepts
1. Frame of Reference
A frame of reference is essential to describe motion. In one-dimensional motion, we choose an origin (x = 0) and a positive direction along the x-axis. The position of the particle at any time t is given by x(t).
Types of frames:
- Inertial frame: A frame in which Newton's first law holds (a body with no net force moves with constant velocity). Example: a frame at rest or moving with constant velocity.
- Non-inertial frame: An accelerating frame. In such frames, pseudo forces must be introduced to apply Newton's laws.
For kinematics (this chapter), any frame works. For dynamics (Chapter 5), only inertial frames are valid.
2. Distance vs. Displacement
| Property | Distance | Displacement | ||
|---|---|---|---|---|
| Definition | Total path length | Shortest path (final − initial position) | ||
| Type | Scalar | Vector | ||
| Can be zero? | Never (for a moving body) | Yes, if object returns to start | ||
| Can be negative? | Never | Yes | ||
| Always ≥? | Distance ≥ | Displacement | Displacement ≤ Distance | |
| Equal when? | Motion in one direction without reversal | Same — also when no reversal | ||
| Example: round trip of 10 m | Distance = 20 m | Displacement = 0 |
Example: A body moves from A to B (5 m east) then B to C (3 m west). Distance = 8 m, Displacement = 2 m (east).
3. Average Speed vs. Average Velocity
- Average speed = Total distance / Total time (always positive)
- Average velocity = Displacement / Total time (can be negative or zero)
Special formulas:
| Condition | Formula | Type of Mean |
|---|---|---|
| Half the distance at v<sub>1</sub>, other half at v<sub>2</sub> | v<sub>avg</sub> = 2v<sub>1</sub>v<sub>2</sub> / (v<sub>1</sub> + v<sub>2</sub>) | Harmonic mean |
| Half the time at v<sub>1</sub>, other half at v<sub>2</sub> | v<sub>avg</sub> = (v<sub>1</sub> + v<sub>2</sub>) / 2 | Arithmetic mean |
| Note: Average speed is always ≥ | Average velocity | . They are equal only for straight-line motion in one direction without reversal. |
|---|
4. Instantaneous Velocity and Speed
Instantaneous velocity v = dx/dt = slope of x-t graph at that point.
| Instantaneous speed = | v | = | dx/dt | = magnitude of instantaneous velocity. |
|---|
If x(t) = 3t<sup>2</sup> + 2t + 5, then v(t) = dx/dt = 6t + 2 and a(t) = dv/dt = 6.
5. Kinematic Equations for Uniformly Accelerated Motion
These three equations relate five variables: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time). They are valid only for constant acceleration.
| Equation | Formula | Variables Involved | When to Use |
|---|---|---|---|
| First equation | v = u + at | u, a, t, v | When time t is known or asked |
| Second equation | s = ut + ½at<sup>2</sup> | u, a, t, s | When displacement is known or asked |
| Third equation | v<sup>2</sup> = u<sup>2</sup> + 2as | u, a, s, v | When time is NOT involved |
| Displacement in nth second | s<sub>n</sub> = u + (a/2)(2n − 1) | u, a, n, s<sub>n</sub> | Distance covered in the nth second specifically |
Derivation of kinematic equations from v-t graph:
- The v-t graph for uniform acceleration is a straight line with slope a.
- v = u + at comes from the equation of the straight line.
- s = ut + ½at<sup>2</sup> comes from the area under the v-t graph (trapezium area).
Sign convention: Take the direction of initial motion as positive. Then:
- Acceleration in the direction of motion → positive a
- Retardation (opposing motion) → negative a
- For free fall: taking downward as positive, a = +g; taking upward as positive, a = −g
6. Velocity-Time and Position-Time Graphs
| Graph | Uniform velocity (a = 0) | Uniform acceleration (a = const, a > 0) | Uniform retardation (a < 0) | Non-uniform acceleration | ||
|---|---|---|---|---|---|---|
| x-t | Straight line (slope = v) | Parabola (curved upward) | Parabola (curved downward) | Curved (shape depends on a(t)) | ||
| v-t | Horizontal line (height = v) | Straight line (slope = a, upward) | Straight line (slope = a, downward) | Curved (slope at a point = a) | ||
| a-t | On time axis (a = 0) | Horizontal line (height = a) | Horizontal line (height = − | a | ) | Varies with time |
What each graph quantity gives:
| Graph Feature | Physical Meaning |
|---|---|
| Slope of x-t graph | Instantaneous velocity |
| Slope of v-t graph | Acceleration |
| Area under v-t graph | Displacement |
| Area under a-t graph | Change in velocity (Δv) |
| x-t graph curvature | Curving up → positive acceleration; curving down → negative acceleration |
Common graph scenarios to practise: 1. Body starts from rest, accelerates uniformly, then moves with constant velocity → v-t: line going up, then horizontal. 2. Body thrown upward, comes back down → v-t: line from positive v, crosses time axis at max height, goes negative. 3. Body decelerates to rest → v-t: line going down to zero; x-t: curve that flattens out.
7. Relative Velocity
- Objects moving in same direction: v<sub>AB</sub> = v<sub>A</sub> − v<sub>B</sub> (magnitude = difference of speeds)
- Objects moving in opposite directions: v<sub>AB</sub> = v<sub>A</sub> + v<sub>B</sub> (magnitude = sum of speeds)
- Object A stationary, B moving: v<sub>AB</sub> = −v<sub>B</sub> (A sees B moving with B's velocity)
- Both at rest or same velocity: v<sub>AB</sub> = 0 (no relative motion)
Rain problem (frequently asked): If rain falls vertically at speed v<sub>r</sub> and a person moves horizontally at speed v<sub>p</sub>:
- The person sees rain at an angle: apparent velocity = √(v<sub>r</sub><sup>2</sup> + v<sub>p</sub><sup>2</sup>)
- Angle from vertical: θ = tan<sup>-1</sup>(v<sub>p</sub>/v<sub>r</sub>)
- Umbrella must be tilted forward at this angle
8. Free Fall Under Gravity
For a body dropped from height h or thrown vertically:
- Replace s with h, a with −g (taking upward as positive) or +g (taking downward as positive).
- v<sup>2</sup> = u<sup>2</sup> + 2gh (with proper sign convention)
- Time to reach ground from height h (dropped, u = 0): t = √(2h/g)
- Maximum height reached when thrown upward with velocity u: H = u<sup>2</sup>/(2g)
- Time to reach max height: t = u/g
- Total time of flight (up and back): T = 2u/g
Symmetry of motion: Under gravity (no air resistance), the upward and downward journeys are symmetric:
- Time to go up = time to come down (from the same point)
- Speed at any height is the same going up and coming down
- The body returns to the starting point with the same speed it was launched
Free fall from height h (key results at a glance):
| Quantity | Dropped (u = 0) | Thrown up (speed u) |
|---|---|---|
| Acceleration | g (downward) | g (downward) |
| Time to reach ground | √(2h/g) | Not a simple formula — solve quadratic |
| Velocity on reaching ground | √(2gh) | √(u<sup>2</sup> + 2gh) |
| Max height reached | h (already at height h) | h + u<sup>2</sup>/(2g) |
Important Numericals
Numerical 1: Basic Kinematics
A car starts from rest and accelerates uniformly at 2 m s<sup>-2</sup> for 10 s. Find the distance covered and final velocity.
Numerical 2: Free Fall
A ball is thrown vertically upward with a velocity of 19.6 m s<sup>-1</sup>. Find the maximum height and total time of flight. (g = 9.8 m s<sup>-2</sup>)
Numerical 3: Relative Velocity
Two trains A and B move on parallel tracks. A moves at 72 km/h and B at 54 km/h in the same direction. Find the relative velocity of A with respect to B. Also, how long does A take to cross B if B is 200 m long?
Numerical 4: Distance in nth Second
A body starts from rest with uniform acceleration of 4 m s<sup>-2</sup>. Find the distance covered in the 5th second.
Numerical 5: Stopping Distance
A car moving at 60 km/h is brought to rest in 5 s. Find the deceleration and stopping distance.
Numerical 6: Two-stage Motion
A body accelerates from rest at 5 m s<sup>-2</sup> for 4 s, then moves with constant velocity for 6 s. Find the total distance.
NCERT Exercise Reference
| NCERT Exercise | Topic | Marks to Expect |
|---|---|---|
| 3.1–3.4 | Frame of reference, displacement, velocity | 1–2 marks (definitions, comparisons) |
| 3.5–3.8 | Kinematic equations | 3–5 marks (numericals) |
| 3.9–3.11 | Graph interpretation | 2–3 marks (sketch or interpret) |
| 3.12–3.14 | Relative velocity | 2 marks (formula + numerical) |
| 3.15–3.18 | Free fall | 2–3 marks (numerical) |
Common Mistakes
1. Confusing distance and displacement: In a round trip, distance is non-zero but displacement is zero. Average velocity = 0, but average speed ≠ 0. 2. Wrong sign convention: Pick one direction as positive and stick to it. If upward is positive, then g = −9.8 m s<sup>-2</sup> for free fall. 3. Using kinematic equations for non-uniform acceleration: The three equations only work for constant a. If a varies, use calculus (integration). 4. Ignoring the "nth second" formula: s<sub>n</sub> = u + (a/2)(2n − 1) gives the distance in the nth second, not total distance in n seconds. Total distance in n seconds = un + ½an<sup>2</sup>. 5. Average speed trap: For two equal distances at different speeds, average speed is the harmonic mean (2v<sub>1</sub>v<sub>2</sub>/(v<sub>1</sub>+v<sub>2</sub>)), not the arithmetic mean. This is a very common MCQ trap. 6. Reading graphs wrong: On a v-t graph, area = displacement (not distance if velocity changes sign); slope = acceleration. On x-t, slope = velocity (not acceleration).
| 7. **Forgetting that instantaneous speed = | instantaneous velocity | **: Speed is always the magnitude of velocity. |
|---|
Exam Tips
- Graph questions are guaranteed: Be prepared to sketch and interpret x-t, v-t, and a-t graphs. Know what each slope and area represents.
- Free fall numericals (2–3 marks): Always state your sign convention before solving. Write "taking upward as positive" or "taking downward as positive."
- Relative velocity is a common 2-mark question: write the formula v<sub>AB</sub> = v<sub>A</sub> − v<sub>B</sub>, substitute, and give the answer with direction.
- For board exams, write the formula first, then substitute values with units. Do not skip steps.
- Displacement in nth second formula appears frequently in 1-mark questions — memorise it.
- When a body is thrown upward, its velocity at the starting point on the way down equals the initial upward velocity (symmetry of motion under gravity, no air resistance).
- Stopping distance problems: Use v<sup>2</sup> = u<sup>2</sup> + 2as with v = 0 and a negative.
- For multi-stage motion problems, solve each stage separately and add distances.
Quick Revision Notes
- Distance = total path length (scalar); Displacement = final − initial position (vector).
- Average speed ≥ |Average velocity|; equal only for straight-line motion in one direction.
- Instantaneous velocity = dx/dt = slope of x-t graph; Instantaneous speed = |v|.
- Kinematic equations (uniform a): v = u + at; s = ut + ½at<sup>2</sup>; v<sup>2</sup> = u<sup>2</sup> + 2as.
- nth second distance: s<sub>n</sub> = u + (a/2)(2n − 1). Derived from s<sub>n</sub> − s<sub>n−1</sub>.
- x-t graph slope = velocity; v-t graph slope = acceleration; v-t graph area = displacement; a-t graph area = Δv.
- Relative velocity: v<sub>AB</sub> = v<sub>A</sub> − v<sub>B</sub>. Same direction → subtract speeds; opposite → add.
- Free fall: a = g downward; max height H = u<sup>2</sup>/2g; time of flight = 2u/g.
- Sign convention: Choose one direction as positive; apply consistently to u, v, a, and s.
- Stopping distance: s = u<sup>2</sup>/(2|a|) when decelerating to rest.
Important One-Mark Facts for Quick Recall
1. In a round trip, displacement is zero but distance is non-zero. Average velocity = 0, average speed ≠ 0. 2. The slope of an x-t graph gives instantaneous velocity. A horizontal x-t graph means the body is at rest. 3. The slope of a v-t graph gives acceleration. A horizontal v-t graph means constant velocity (a = 0). 4. The area under a v-t graph gives displacement. Area above the time axis is positive; below is negative. 5. The area under an a-t graph gives change in velocity (Δv), not displacement. 6. For two equal distances at speeds v<sub>1</sub> and v<sub>2</sub>, average speed = 2v<sub>1</sub>v<sub>2</sub>/(v<sub>1</sub>+v<sub>2</sub>) (harmonic mean, NOT arithmetic mean). 7. A body in free fall has a = g at all points, regardless of whether it is going up or down. 8. At the highest point of a vertically thrown body, v = 0 but a = g (acceleration does not become zero). 9. The kinematic equations are valid only for constant acceleration. 10. If a body covers equal displacements in equal time intervals, it has zero acceleration (uniform velocity). 11. The direction of acceleration is the same as the direction of the net force, not necessarily the direction of velocity. 12. For a body starting from rest (u = 0), distance ∝ t<sup>2</sup> (s = ½at<sup>2</sup>) — this means distances in successive seconds are in the ratio 1:3:5:7...
Assertion-Reason Practice
| Assertion | Reason | Verdict | ||
|---|---|---|---|---|
| A body can have zero average velocity but non-zero average speed. | Average velocity depends on displacement; average speed depends on distance. | A is true, R is the correct explanation. A round trip has zero displacement but non-zero distance. | ||
| A body can have constant speed but varying velocity. | Velocity is a vector; its direction can change while magnitude stays constant. | A is true, R is the correct explanation. Uniform circular motion is the example. | ||
| The distance covered in the nth second equals the total distance in n seconds. | s_n = u + (a/2)(2n − 1). | A is false, R is true. s_n gives only the distance in the nth second, not total distance. | ||
| At the highest point of vertical motion, acceleration is zero. | Velocity at the highest point is zero. | A is false, R is true. Acceleration = g throughout, even when v = 0 at the top. | ||
| Average speed is always greater than or equal to average velocity. | Distance ≥ | Displacement | . | A is true, R is the correct explanation. |
How to Interpret Graphs — Step-by-Step
Graph interpretation is the most common exam skill tested from this chapter. Follow this systematic approach:
Step 1: Identify the graph type — x-t, v-t, or a-t.
Step 2: Check the slope:
- x-t graph slope = velocity. Upward slope = positive velocity, downward = negative, zero slope = at rest.
- v-t graph slope = acceleration. Upward = speeding up, downward = slowing down, zero = constant velocity.
- a-t graph slope = rate of change of acceleration (jerk, not in syllabus).
Step 3: Check the curvature (x-t graph only):
- Curving upward (concave up) = positive acceleration.
- Curving downward (concave down) = negative acceleration.
- Straight line = zero acceleration.
Step 4: Check the area (v-t and a-t graphs):
- v-t graph area = displacement. Count area above time axis as positive, below as negative.
- a-t graph area = change in velocity (Δv).
Step 5: Identify special points:
- Where x-t graph slope = 0 → body at rest.
- Where v-t graph crosses time axis → body momentarily at rest (turning point).
- Where v-t graph slope = 0 → constant velocity (zero acceleration).