NCERT Class 11 Physics • Chapter 2 Reprint 2026-27

Motion in a Straight Line

Complete chapter notes, calculus & graphical derivations, stepwise worked examples 2.1–2.7, concept check quiz with option explanations, textbook exercises 2.1–2.18, and summary reference.
01 / Chapter Theory

Motion in a Straight Line

Official NCERT syllabus: position, path length & displacement, instantaneous velocity & speed, acceleration, kinematic equations by graphical and calculus methods, free fall, stopping distance, reaction time, and relative velocity.

5 SECTIONS
Chapter Navigation
2.1Introduction 2.2Instantaneous Velocity & Speed 2.3Acceleration 2.4Kinematic Equations of Motion 2.5Relative Velocity
2.1
NCERT Section

Introduction

Fundamental Concept

2.1.1 Motion & Kinematics Scope

Motion is the change in position of an object with time. Kinematics is the branch of mechanics that describes motion without going into the causes of motion (forces, which are studied in Chapter 4).

  • Rectilinear Motion: Motion of an object along a straight line.
  • Point Object Approximation: An object is treated as a point-like object when its size is much smaller than the distance it traverses in a reasonable duration of time (e.g. an aeroplane flying between cities, the Earth revolving around the Sun).
  • Frame of Reference: A coordinate system with a clock attached, anchored to a chosen origin O. Position to the right of the origin is taken as positive (+), and to the left as negative (−).
Path Length vs. Displacement: Path length is the total length of the actual path traversed (a scalar quantity ≥ 0). Displacement (Δx = x2 − x1) is the shortest directed vector from the initial to the final position. Magnitude of displacement ≤ Path length.
2.2
NCERT Section

Instantaneous Velocity and Speed

Calculus Definition

2.2.1 Instantaneous Velocity

While average velocity Δx/Δt describes motion over an extended interval, instantaneous velocity (v) describes how fast an object moves at a specific instant t.

It is defined mathematically as the limit of the average velocity as the time interval Δt approaches zero:

v = limΔt → 0 (Δx / Δt) = dx / dt
(The first derivative of position with respect to time)

Graphical Meaning: The instantaneous velocity at any time t equals the slope of the tangent to the position-time (x-t) curve at that point.

Time t (s) Position x (m) P₁ P₂ P (t = 4.0 s) Tangent Slope = v = dx/dt t = 3 s t = 4 s t = 5 s
Figure 2.1: Determining instantaneous velocity from an x-t graph. As Δt → 0, the chord P1P2 becomes the tangent at point P, whose slope equals v = dx/dt.
Interactive 3D Position-Time Curve & Instantaneous Tangent Slope ($v = dx/dt$)
Calculus Lab
Drag to inspect tangent in 3D
Instantaneous Velocity & Tangent Slope
Time: t = 2.0s • Position: x(t) = 2.0m • Tangent Slope: v = dx/dt = 2.00 m/s
Calculus Principle: The slope of the rose tangent line at any coordinate point on the position-time curve represents the exact instantaneous velocity $v(t) = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$.
Table 2.1 • NCERT Reference

2.2.2 Limiting Value of Δx/Δt at t = 4.0 s (for x = 0.08 t3)

Δt (s) t1 (s) t2 (s) x(t1) (m) x(t2) (m) Δx (m) Δx/Δt (m s−1)
2.03.05.02.1610.007.843.92
1.03.54.53.437.293.863.86
0.53.754.254.218756.141251.92253.845
0.13.954.054.930395.314410.384023.8402
0.013.9954.0055.1008245.1392240.038403.8400

As Δt → 0, the average velocity converges to the exact derivative value: v(4.0 s) = 3.84 m s−1 (since dx/dt = 0.24 t2 = 0.24 × 16 = 3.84).

Instantaneous Speed vs. Instantaneous Velocity: Instantaneous speed is the pure magnitude of instantaneous velocity (|v|). A velocity of +24.0 m s−1 and −24.0 m s−1 both have an instantaneous speed of 24.0 m s−1.

Example 2.1 • Calculus Velocity from Position Equation

NCERT Solved

The position of an object moving along the x-axis is given by x = a + bt2 where a = 8.5 m, b = 2.5 m s−2, and t is in seconds. What is its velocity at t = 0 s and t = 2.0 s? What is the average velocity between t = 2.0 s and t = 4.0 s?

Final Results: v(0 s) = 0 m s−1  |  v(2.0 s) = 10.0 m s−1  |  vavg(2 to 4 s) = 15.0 m s−1
Step 1: Differentiate position function to find velocity.
v(t) = dx/dt = d/dt (a + bt2) = 2bt = 2(2.5)t = 5.0 t m s−1.
Step 2: Calculate instantaneous velocities.
At t = 0 s: v(0) = 5.0(0) = 0 m s−1.
At t = 2.0 s: v(2.0) = 5.0(2.0) = 10.0 m s−1.
Step 3: Calculate average velocity between 2.0 s and 4.0 s.
x(4.0 s) = 8.5 + 2.5(4.0)2 = 8.5 + 40.0 = 48.5 m.
x(2.0 s) = 8.5 + 2.5(2.0)2 = 8.5 + 10.0 = 18.5 m.
vavg = [x(4.0) − x(2.0)] / [4.0 − 2.0] = (48.5 − 18.5) / 2.0 = 30.0 / 2.0 = 15.0 m s−1.
2.3
NCERT Section

Acceleration

Rate of Change of Velocity

2.3.1 Definition & Graphs

Acceleration is the rate of change of velocity with time.

  • Average Acceleration: aavg = (v2 − v1) / (t2 − t1) = Δv / Δt. SI unit: m s−2.
  • Instantaneous Acceleration: a = limΔt → 0 (Δv / Δt) = dv / dt = d2x / dt2 (slope of tangent on v-t graph).
(a) a > 0 (Curves Up) (b) a < 0 (Curves Down) (c) a = 0 (Straight Line)
Figure 2.2: Position-time (x-t) graphs for motion with (a) positive acceleration (concave up), (b) negative acceleration (concave down), and (c) zero acceleration (uniform velocity).
(a) v > 0, a > 0 Speeding Up (+x) (b) v > 0, a < 0 Slowing Down (+x) (c) v < 0, a < 0 Speeding Up (−x) t₁ (d) Reverses at t₁ Turns Back
Figure 2.3: Velocity-time graphs for constant acceleration: (a) positive direction speeding up, (b) positive direction slowing down, (c) negative direction speeding up, (d) moving in positive direction till t1 then turning back.
Time t Velocity v u T Area = Height × Base = u × T = Displacement (x)
Figure 2.4: The area under the velocity-time (v-t) curve between t = 0 and t = T equals the displacement x = uT.
2.4
NCERT Section

Kinematic Equations for Uniformly Accelerated Motion

Derivation & Master Formulae

2.4.1 Derivation of the Three Kinematic Equations

For motion with constant acceleration (a), we derive the three fundamental relations connecting initial velocity v0, final velocity v, acceleration a, time t, and displacement x:

v₀ (A) v (B) O t (D) C Triangle Area = ½ (v − v₀) t = ½ a t² Rectangle Area = v₀ t
Figure 2.5: Total area under v-t curve = Area(ΔABC) + Area(OACD) = v0t + ½at2.
Kinematic Master Equations (Constant Acceleration $a$) $$v = v_0 + at$$ $$x = v_0 t + \frac{1}{2} a t^2 \quad \left(\text{or } x = x_0 + v_0 t + \frac{1}{2} a t^2\right)$$ $$v^2 = v_0^2 + 2ax \quad \left(\text{or } v^2 = v_0^2 + 2a(x - x_0)\right)$$ $$\bar{v} = \frac{v_0 + v}{2} \implies x = \left(\frac{v_0 + v}{2}\right) t$$
Interactive 3D 1D Uniform Accelerated Motion & Kinematics
Kinematics Lab
Drag to change perspective
Instantaneous Kinematic Parameters
Elapsed Time: t = 0.0s • Speed: v = 2.0 m/s • Position: x = 0.0 m
Kinematic Formulations: Speed grows linearly according to $v(t) = v_0 + at$, while position increases quadratically as $x(t) = v_0 t + \frac{1}{2}at^2$.

Example 2.2 • Calculus Derivation of Equations of Motion

NCERT Solved

Obtain equations of motion for constant acceleration using the method of calculus.

Step 1: First Equation ($v = v_0 + at$): $$a = \frac{dv}{dt} \implies dv = a\, dt$$ $$\int_{v_0}^v dv = a \int_0^t dt \implies [v - v_0] = at \implies v = v_0 + at$$
Step 2: Second Equation ($x = x_0 + v_0 t + \frac{1}{2}at^2$): $$v = \frac{dx}{dt} \implies dx = v\, dt = (v_0 + at)\, dt$$ $$\int_{x_0}^x dx = \int_0^t (v_0 + at)\, dt \implies x - x_0 = v_0 t + \frac{1}{2}at^2 \implies x = x_0 + v_0 t + \frac{1}{2}at^2$$
Step 3: Third Equation ($v^2 = v_0^2 + 2a(x - x_0)$): $$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} \implies v\, dv = a\, dx$$ $$\int_{v_0}^v v\, dv = a \int_{x_0}^x dx \implies \frac{1}{2}(v^2 - v_0^2) = a(x - x_0) \implies v^2 = v_0^2 + 2a(x - x_0)$$

Example 2.3 • Ball Thrown Upwards from a Building

NCERT Solved

A ball is thrown vertically upwards with a velocity of $20\text{ m s}^{-1}$ from the top of a multistorey building $25.0\text{ m}$ high from the ground. (a) How high will the ball rise? (b) How long will it be before the ball hits the ground? (Take $g = 10\text{ m s}^{-2}$).

Answers: (a) Rise height $= 20.0\text{ m}$ ($45.0\text{ m}$ from ground)  |  (b) Total flight time $= 5.0\text{ s}$
Ground (y = 0) A (y₀ = 25 m) B: Peak (v = 0, y = 45 m) t₁ = 2.0 s C: Ground (t = 5.0 s) +y (Upward) a = −g = −10 m s⁻² v₀ = +20 m s⁻¹
Figure 2.6: Coordinate framework for Example 2.3: Ball launched from top of building (A) rising to peak (B) and striking ground (C).
(a) Maximum height rise: At peak, $v = 0$. Using $v^2 = v_0^2 + 2a(y - y_0)$ with $a = -10\text{ m s}^{-2}$: $$0 = (20)^2 + 2(-10)(y - y_0) \implies 20(y - y_0) = 400 \implies y - y_0 = 20.0\text{ m}$$ Total height from ground $= 25.0\text{ m} + 20.0\text{ m} = 45.0\text{ m}$.
(b) Total flight time to ground (Quadratic formulation): Taking launch point $y_0 = 0$, ground is at $y = -25.0\text{ m}$, $v_0 = +20\text{ m s}^{-1}$, $a = -10\text{ m s}^{-2}$: $$y = v_0 t + \frac{1}{2} a t^2 \implies -25 = 20t - 5t^2 \implies 5t^2 - 20t - 25 = 0 \implies t^2 - 4t - 5 = 0$$ $$(t - 5)(t + 1) = 0 \implies t = 5.0\text{ s} \quad (\text{rejecting negative time } t = -1\text{ s})$$
Interactive 3D Ball Throw from Building & Free Fall (Example 2.3)
Gravity Lab
Drag to change perspective
Live Trajectory State & Time Breakdown
t = 0.00s • Height from Ground: y = 25.0m • Velocity: v = +20.0 m/s • Ascending to Peak
Launch from Building (25m) at +20 m/s → Peaks at 45m (t=2s) → Hits ground at t=5s

Example 2.4 • Free Fall under Gravity

NCERT Solved

Discuss the motion of an object under free fall. Neglect air resistance and state equations for velocity, displacement, and acceleration.

−9.8 (a) a = −9.8 m s⁻² (b) v = −9.8 t (c) y = −4.9 t²
Figure 2.7: Motion under free fall: (a) Acceleration-time, (b) Velocity-time, (c) Distance-time variations with time.
Governing Equations (taking upward as +y, released from rest v0 = 0):
• Acceleration: a = −g = −9.8 m s−2 (constant).
• Velocity: v = −gt = −9.8 t m s−1.
• Position: y = −½ gt2 = −4.9 t2 m.
• Velocity-Position: v2 = −2gy = −19.6 y m2 s−2.

Example 2.5 • Galileo's Law of Odd Numbers

NCERT Solved

Prove Galileo's law of odd numbers: “The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity (1 : 3 : 5 : 7 : ...).”

Ratio of distances in successive equal intervals τ = 1 : 3 : 5 : 7 : 9 : 11...
Time t Total Position y In units of y0 (½gτ2) Distance in interval Δy Ratio of Distances
000--
τ−½ gτ2y0y01
2τ−½ g(2τ)2 = −4(½gτ2)4y03y03
3τ−½ g(3τ)2 = −9(½gτ2)9y05y05
4τ−½ g(4τ)2 = −16(½gτ2)16y07y07
5τ−½ g(5τ)2 = −25(½gτ2)25y09y09

Example 2.6 • Stopping Distance of Vehicles

NCERT Solved

When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. Derive an expression for stopping distance ds in terms of initial velocity v0 and deceleration −a.

Formula: ds = v02 / (2a) → ds ∝ v02
Step 1: Apply third equation of motion.
Final velocity v = 0, acceleration = −a. Using v2 = v02 + 2(−a)ds → 0 = v02 − 2a ds.
Step 2: Solve for stopping distance.
ds = v02 / 2a.
Doubling the initial speed increases stopping distance by a factor of 4 (22).

Example 2.7 • Reaction Time Measurement

NCERT Solved

Reaction time is the time taken to observe, think, and act. A ruler dropped vertically between a student's fingers falls through a distance d = 21.0 cm before being caught. Estimate the reaction time (g = 9.8 m s−2).

Answer: Reaction Time tr ≈ 0.21 s (210 ms)
0 cm d = 21 cm Caught by finger d = 21.0 cm Reaction Time Formula: tᵣ = √(2d / g) tᵣ = √(0.42 / 9.8) ≈ 0.21 s
Figure 2.8: Measuring reaction time using a dropped ruler: tr = √(2d/g).
Step 1: Set up free fall distance formula.
Since v0 = 0, d = ½ g tr2 → tr = √(2d / g).
Step 2: Substitute given values (d = 0.21 m, g = 9.8 m s−2).
tr = √[2(0.21) / 9.8] = √(0.42 / 9.8) = √(0.042857) ≈ 0.21 s.
2.5
NCERT Section

Relative Velocity

Relative Motion in 1D

2.5.1 Relative Velocity Formula

The velocity of object B relative to object A is given by:

vBA = vB − vA  |  vAB = vA − vB = −vBA
  • Same Direction: Relative speed = |vA − vB| (subtraction).
  • Opposite Direction: Relative speed = vA − (−vB) = vA + vB (addition).
Interactive 3D 1D Relative Motion & Overtaking Simulator
Relative Motion Lab
Drag to change perspective
Relative Motion Velocity Computation
Car A: 15 m/s | Car B: 10 m/s (Same) ➔ Relative Velocity: v_AB = v_A − v_B = 5.0 m/s
Relative Velocity Takeaway: The relative velocity of $A$ with respect to $B$ is $\mathbf{v}_\text{AB} = \mathbf{v}_\text{A} - \mathbf{v}_\text{B}$. When moving in opposite directions, the minus signs combine: $v_\text{AB} = v_\text{A} - (-v_\text{B}) = v_\text{A} + v_\text{B}$.
02 / Self-Assessment Lab

Concept Check Questions

Select any option to reveal stepwise explanations for every individual choice, detailing why the correct answer is valid and why the alternatives are incorrect.

20 QUESTIONS
Progress: 0 / 20 Answered
Score: 0
Q01 UNANSWERED

The slope of the tangent to a position-time (x-t) graph at any instant represents:

Option A is incorrect. Average acceleration is Δv/Δt, which is obtained from a velocity-time graph chord.
Option B is correct. By calculus definition, v = dx/dt, which is the slope of the tangent to the x-t curve.
Option C is incorrect. Path length is the total distance traversed, not a derivative.
Option D is incorrect. Instantaneous acceleration is the slope of the v-t graph, or second derivative d2x/dt2.
Core Rule
Slope of x-t graph = Velocity (dx/dt). Slope of v-t graph = Acceleration (dv/dt).
Q02 UNANSWERED

The area under a velocity-time (v-t) graph represents:

Option A is incorrect. Instantaneous acceleration is the slope of the v-t graph.
Option B is correct. ∫ v dt = ∫ (dx/dt) dt = Δx (Displacement). Dimensionally, [m s−1] × [s] = [m].
Option C is incorrect. Average speed is total path length divided by total time.
Option D is incorrect. Rate of change of acceleration is jerk (da/dt).
Core Rule
Area under v-t curve = Displacement. Area under a-t curve = Change in velocity.
Q03 UNANSWERED

Can a particle have zero velocity at an instant and yet have non-zero acceleration?

Option A is correct. When a ball reaches its highest point, its instantaneous velocity is momentarily zero (v = 0), but it experiences continuous downward acceleration g = 9.8 m s−2.
Option B is incorrect. Acceleration is the rate of change of velocity, which can be non-zero even when v crosses zero.
Option C is incorrect. It occurs routinely in 1D rectilinear vertical projection.
Option D is incorrect. Derivatives can be non-zero at points where the function value itself is zero (e.g. d/dt(t) = 1 at t=0).
Core Rule
Zero velocity does not imply zero acceleration (e.g. top of projectile trajectory).
Q04 UNANSWERED

If the speed of a vehicle is doubled, its stopping distance (for the same constant braking deceleration) becomes:

Option A is incorrect. Stopping distance is proportional to the square of initial velocity, not linear.
Option B is correct. From v2 = v02 − 2ads → ds = v02 / 2a. Doubling v0 quadruples ds (22 = 4).
Option C is incorrect. 8 times would correspond to a cubic power dependency.
Option D is incorrect. Kinetic energy (½mv2) that must be dissipated by friction scales as v2.
Core Rule
Stopping distance ds ∝ v02 (proportional to the square of initial velocity).
Q05 UNANSWERED

According to Galileo's law of odd numbers, the distances fallen by a body from rest in successive equal time intervals are in the ratio:

Option A is incorrect. 1 : 2 : 3 : 4 is the ratio of velocities (v = gt), not interval distances.
Option B is correct. Total distance at nτ is ½g(nτ)2 ∝ n2 (1, 4, 9, 16, 25). Differences between successive squares give the odd numbers: (1−0), (4−1)=3, (9−4)=5, (16−9)=7...
Option C is incorrect. 1 : 4 : 9 : 16 is the ratio of TOTAL cumulative distances from release, not individual successive interval distances.
Option D is incorrect. Equal distances occur only in uniform motion without acceleration.
Core Rule
In free fall from rest: Cumulative distance ∝ t2 (1, 4, 9, 16). Interval distance ∝ Odd numbers (1, 3, 5, 7).
Q06 UNANSWERED

Two trains A and B are moving in opposite directions along parallel tracks with speeds 72 km h−1 and 54 km h−1 respectively. The relative speed of train B with respect to A is:

Option A is incorrect. 18 km h−1 is the difference, which applies only when moving in the SAME direction.
Option B is correct. In opposite directions: vBA = vB − (−vA) = 54 − (−72) = 126 km h−1 = 126 × (5/18) = 35 m s−1.
Option C is incorrect. Speed of A alone.
Option D is incorrect. Relative speed is zero only when moving with equal velocities in the same direction.
Core Rule
When moving in opposite directions, relative speed is the sum of individual speeds (vrel = v1 + v2).
Q07 UNANSWERED

For a particle with x = 8.5 + 2.5 t2 (in metres), the acceleration of the particle is:

Option A is incorrect. 2.5 is the coefficient b; acceleration is 2b.
Option B is correct. v = dx/dt = 5.0t. Acceleration a = dv/dt = d/dt(5.0t) = 5.0 m s−2 (constant).
Option C is incorrect. 8.5 is the initial position coordinate x0.
Option D is incorrect. The position function is quadratic in time, meaning acceleration is constant and non-zero.
Core Rule
Comparing x = x0 + ½at2 with x = 8.5 + 2.5t2 gives ½a = 2.5 → a = 5.0 m s−2.
Q08 UNANSWERED

A ball thrown vertically upwards rises to a maximum height h. The ratio of time taken to go up (tup) to time taken to fall back (tdown) neglecting air resistance is:

Option A is correct. Under constant gravitational acceleration −g, ascending time tup = v0/g equals descending time tdown = √(2h/g) = v0/g.
Option B is incorrect. Ascent and descent are symmetrical under uniform gravity without air drag.
Option C is incorrect. Upward and downward flight times are equal.
Option D is incorrect. Free fall acceleration g is independent of mass.
Core Rule
Time of ascent = Time of descent = v0/g for any body projected under uniform gravity without drag.
Q09 UNANSWERED

Which of the following is true for an object in rectilinear motion with negative acceleration?

Option A is incorrect. If the object is moving in the negative direction (v < 0) with a < 0, its speed |v| INCREASES.
Option B is incorrect. A straight line on x-t indicates zero acceleration.
Option C is correct. Since d2x/dt2 = a < 0, the second derivative is negative, creating downward curvature.
Option D is incorrect. Velocity can be positive, zero, or negative under negative acceleration.
Core Rule
Sign of acceleration dictates graph curvature (a > 0 curves up, a < 0 curves down). Speed increases when v and a have matching signs.
Q10 UNANSWERED

Under what condition is the magnitude of average velocity equal to average speed?

Option A is not sufficient. A body can move at constant speed in a loop, giving zero average velocity.
Option B is correct. Only when motion is strictly unidirectional along a straight line does path length equal magnitude of displacement, making average speed equal magnitude of average velocity.
Option C is incorrect. Returning to starting position gives average velocity = 0, but average speed > 0.
Option D is incorrect. In general, average speed ≥ |average velocity|.
Core Rule
Average speed = |Average velocity| if and only if the object travels along a straight line in a single unchanging direction.
Q11 UNANSWERED

Why is instantaneous speed always strictly equal to the magnitude of instantaneous velocity?

Option A is incorrect. Instantaneous acceleration does not need to be zero.
Option B is correct. For an infinitesimally small time interval dt, the particle has no time to reverse direction or curve; hence ds = |dx|, making instantaneous speed ds/dt = |dx/dt| = |v|.
Option C is incorrect. Speed is a scalar quantity.
Option D is incorrect. Velocity has dimensions [L T−1].
Core Rule
In the infinitesimal limit Δt → 0, arc length Δs and chord |Δx| become identical.
Q12 UNANSWERED

A police van moving at 30 km h−1 fires a bullet with muzzle speed 150 m s−1 at a thief's car speeding away in the same direction at 192 km h−1. The impact speed of the bullet is:

Option A is incorrect. Muzzle speed is relative to the van, ignoring vehicle motions.
Option B is correct. vvan = 30 × 5/18 = 8.33 m s−1; vbullet,ground = 150 + 8.33 = 158.33 m s−1. vthief = 192 × 5/18 = 53.33 m s−1. Relative impact speed = 158.33 − 53.33 = 105 m s−1.
Option C is incorrect. Adds the thief's speed instead of subtracting.
Option D is incorrect. Subtracted 150 from 192 directly without converting units.
Core Rule
Effective impact speed = (Muzzle speed + Van speed) − Target speed (all in matching units m s−1).
Q13 UNANSWERED

Which of the following cannot possibly represent one-dimensional motion of a particle?

Option A is correct. A single object cannot occupy two different positions or possess two different velocities at the exact same instant of time.
Option B is physically valid. Negative slope simply indicates velocity in the negative direction.
Option C is physically valid. Crossing the time axis represents a particle stopping and reversing direction.
Option D is physically valid. A parabola in x-t represents uniform acceleration (x = v0t + ½at2).
Core Rule
Time is strictly monotonic and one-way; multi-valued functions of time are physically impossible.
Q14 UNANSWERED

A ball dropped from height 90 m on a floor loses 1/10 of its speed on each bounce. The speed with which it rebounds after the first bounce is:

Option A is correct. Impact speed v = √(2gh) = √(2 × 9.8 × 90) = √1764 = 42 m s−1. Losing 1/10 speed leaves 9/10: vrebound = 0.9 × 42 = 37.8 m s−1.
Option B is incorrect. 42 m s−1 is the speed before impact.
Option C is incorrect. 4.2 m s−1 is the loss in speed, not the retained rebound speed.
Option D is incorrect. Incorrect calculation.
Core Rule
Impact speed v = √(2gh). Rebound speed = (1 − fractional loss) × v.
Q15 UNANSWERED

A drunkard takes 5 steps forward (1 m each, 1 s each) and 3 steps backward, repeating continuously. How long does he take to fall into a pit 13 m away?

Option A is incorrect. Assumes he must complete full backward cycles even after falling into the pit!
Option B is correct. In 1 cycle of 8 s, net forward displacement = 5 − 3 = 2 m. In 4 cycles (32 s), net distance = 8 m. On the 5th forward leg, taking 5 steps (5 s) reaches 8 + 5 = 13 m (into the pit!). Total time = 32 + 5 = 37 s.
Option C is incorrect. Fails to account for exact landing moment on forward step 5.
Option D is incorrect. Only true if he never stepped backward.
Core Rule
When an absorbing boundary (pit) is reached during the forward motion, subsequent backward steps never occur.
Q16 UNANSWERED

A car moving at 126 km h−1 stops in 200 m. The uniform retardation and stopping time are respectively:

Option A is correct. v0 = 126 × (5/18) = 35 m s−1. Using 0 = (35)2 − 2a(200) → a = 1225 / 400 = 3.06 m s−2. Time t = v0/a = 35 / 3.0625 = 11.43 s.
Option B is incorrect. Incorrect calculation.
Option C is incorrect. Retardation is 3.06 m s−2.
Option D is incorrect. Stopping time is 11.4 s, not 22.8 s.
Core Rule
Convert speed to SI (35 m s−1), apply v2 = v02 + 2ax, then v = v0 + at.
Q17 UNANSWERED

In a simple harmonic motion x-t curve (x = A sin ωt), the acceleration is always:

Option A is correct. In SHM, restoring acceleration is directed towards the center (a = −ω2x). When x > 0, a < 0; when x < 0, a > 0.
Option B is incorrect. Acceleration alternates signs periodically.
Option C is incorrect. When moving away from origin, acceleration opposes velocity.
Option D is incorrect. Acceleration varies continuously with displacement x.
Core Rule
In harmonic motion, a(t) = −ω2x(t); acceleration is proportional and opposite to displacement.
Q18 UNANSWERED

A man walks 2.5 km to market at 5 km h−1, finds it closed, and immediately walks back at 7.5 km h−1. His average speed for the whole round trip (0 to 50 min) is:

Option A is incorrect. Zero is the magnitude of average velocity, NOT average speed.
Option B is correct. Outward time = 2.5/5 = 0.5 h (30 min). Return time = 2.5/7.5 = 1/3 h (20 min). Total path length = 2.5 + 2.5 = 5.0 km. Total time = 0.5 + 1/3 = 5/6 h (50 min). Average speed = 5.0 / (5/6) = 6.0 km h−1.
Option C is incorrect. 6.25 is arithmetic average (5+7.5)/2, which is invalid because time intervals differ.
Option D is incorrect. Outward speed alone.
Core Rule
Average speed = Total Path Length / Total Time = 2v1v2 / (v1 + v2) for equal distance legs.
Q19 UNANSWERED

At turning points or peak vertices (A, B, C) on a speed-time graph, the instantaneous acceleration is:

Option A is correct. At smooth extrema/peaks of a speed-time curve, the tangent is horizontal (slope = dv/dt = 0), hence acceleration is zero.
Option B is incorrect. Maximum acceleration occurs where the slope is steepest, not at horizontal turning points.
Option C is incorrect. Infinite acceleration would require a vertical jump.
Option D is incorrect. Slope is zero at the local stationary point.
Core Rule
Extrema (maxima/minima) of smooth v-t curves have zero tangent slope → a = 0.
Q20 UNANSWERED

The kinematic equations v = v0 + at and x = v0t + ½at2 are strictly applicable ONLY when:

Option A is incorrect. Variable acceleration requires integration of a(t).
Option B is correct. These algebraic formulas assume a = constant during integration. For variable acceleration, calculus definitions v = dx/dt and a = dv/dt must be integrated directly.
Option C is incorrect. If speed is constant, a = 0 (a special trivial case).
Option D is incorrect. Circular motion involves continuously changing acceleration direction.
Core Rule
Standard kinematic equations hold exclusively for uniform (constant) acceleration along a line.
03 / Textbook Solutions

NCERT Exercises 2.1 – 2.18

Complete stepwise solutions for every textbook exercise question in NCERT Class 11 Physics Chapter 2 Reprint 2026-27.

18 QUESTIONS
Ex 2.1 Point Object Approximation

2.1 In which of the following examples of motion can the body be considered approximately a point object:

(a)
A railway carriage moving without jerks between two stations.
(b)
A monkey sitting on top of a man cycling smoothly on a circular track.
(c)
A spinning cricket ball that turns sharply on hitting the ground.
(d)
A tumbling beaker that has slipped off the edge of a table.
Prerequisite: ↗ 2.1 Point Object Approximation
Summary: (a) and (b) are point objects; (c) and (d) cannot be treated as point objects.
(a) Railway carriage:
Yes. The distance between two railway stations is vastly greater than the dimensions of the carriage.
(b) Monkey on cyclist on circular track:
Yes. The size of the monkey is negligible compared to the length of the cycling track.
(c) Spinning cricket ball turning sharply:
No. The spinning rotation and sharp turning trajectory depend on the finite size and rotational dynamics of the ball.
(d) Tumbling beaker off table edge:
No. The height of the table is comparable to the size of the beaker, and its tumbling motion involves rotational orientation.
Ex 2.2 Position-Time Graph Analysis

2.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below:

t x O P Q A B
Figure 2.9: x-t graph for children A and B returning from school O to homes P and Q.
(a)
(A/B) lives closer to the school than (B/A)
(b)
(A/B) starts from the school earlier than (B/A)
(c)
(A/B) walks faster than (B/A)
(d)
A and B reach home at the (same/different) time
(e)
(A/B) overtakes (B/A) on the road (once/twice).
Answers: (a) A lives closer than B  |  (b) A starts earlier than B  |  (c) B walks faster than A  |  (d) same time  |  (e) B overtakes A once.
(a) Proximity: OP < OQ, so A lives closer to school than B.
(b) Start time: A starts at t = 0, while B starts at t > 0. Thus A starts earlier.
(c) Speed (Slope): The slope of B's line is steeper than A's line, so B walks faster than A.
(d) Arrival time: Dropping verticals from the endpoints of both graphs to the time axis shows both reach their respective homes at the same time t.
(e) Overtaking: The two graphs intersect once; therefore B overtakes A once on the road.
Ex 2.3 x-t Graph Construction

2.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h−1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h−1. Choose suitable scales and plot the x-t graph of her motion.

Timeline: Walks 9:00 to 9:30 am • Stays at office 9:30 am to 5:00 pm • Auto return 5:00 to 5:06 pm
Step 1: Calculate walk time to office.
Time = Distance / Speed = 2.5 km / 5 km h−1 = 0.5 h = 30 minutes (9:00 am → 9:30 am).
Step 2: Office stay.
From 9:30 am to 5:00 pm, x = 2.5 km (stationary, horizontal line).
Step 3: Return trip by auto.
Time = 2.5 km / 25 km h−1 = 0.1 h = 6 minutes (5:00 pm → 5:06 pm).
Ex 2.4 Periodic Step Motion

2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Answer: Total time = 37 s
Cycle analysis:
1 cycle = 5 s forward (5 m) + 3 s backward (3 m) = 8 s time, covering net displacement 2 m.
In 4 cycles (4 × 8 s = 32 s), net displacement = 4 × 2 m = 8 m.
Remaining distance to pit = 13 m − 8 m = 5 m.
On the 5th cycle, he moves 5 m forward in 5 s, landing in the pit at x = 13 m.
Total time = 32 s + 5 s = 37 s.
Ex 2.5 Retardation & Stopping Time

2.5 A car moving along a straight highway with a speed of 126 km h−1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Answers: Retardation a = 3.06 m s−2  |  Stopping time t = 11.4 s
Step 1: Convert units to SI.
v0 = 126 × (1000 / 3600) = 35 m s−1, v = 0, x = 200 m.
Step 2: Calculate retardation.
v2 = v02 + 2ax → 0 = (35)2 + 2a(200) → 400a = −1225 → a = −3.06 m s−2 (Retardation = 3.06 m s−2).
Step 3: Calculate time to stop.
v = v0 + at → 0 = 35 − 3.0625t → t = 35 / 3.0625 = 11.43 s ≈ 11.4 s.
Ex 2.6 Vertical Projection

2.6 A player throws a ball upwards with an initial speed of 29.4 m s−1:

(a)
What is the direction of acceleration during the upward motion of the ball?
(b)
What are the velocity and acceleration of the ball at the highest point of its motion?
(c)
Choose x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be positive, and give the signs of position, velocity, and acceleration during upward and downward motion.
(d)
To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s−2).
Answers: (a) Vertically downward • (b) v = 0, a = 9.8 m s−2 downward • (d) Height = 44.1 m, Total time = 6.0 s
(a) Direction of acceleration: Vertically downward at all times due to gravity.
(b) At highest point: Velocity v = 0, acceleration a = 9.8 m s−2 (downwards).
(c) Signs with origin at highest point (+ downward):
• Upward motion: Position x > 0 (+), Velocity v < 0 (−), Acceleration a > 0 (+).
• Downward motion: Position x > 0 (+), Velocity v > 0 (+), Acceleration a > 0 (+).
(d) Maximum height and total time:
h = v02 / (2g) = (29.4)2 / (2 × 9.8) = 864.36 / 19.6 = 44.1 m.
Time of ascent = v0/g = 29.4 / 9.8 = 3.0 s. Total return time = 2 × 3.0 = 6.0 s.
Ex 2.7 True / False Conceptual Analysis

2.7 State with reasons if each statement is True or False for a particle in one-dimensional motion:

(a)
With zero speed at an instant may have non-zero acceleration at that instant.
(b)
With zero speed may have non-zero velocity.
(c)
With constant speed must have zero acceleration.
(d)
With positive value of acceleration must be speeding up.
(a) True: A ball at the highest point of its vertical trajectory has speed = 0, but acceleration = g = 9.8 m s−2 downwards.
(b) False: Speed is the magnitude of velocity. If speed = 0, magnitude of velocity = 0, hence velocity must be zero.
(c) True (in 1D motion): In strictly one-dimensional rectilinear motion, constant speed implies velocity cannot change sign without stopping, hence acceleration dv/dt = 0.
(d) False: If the particle is moving in the negative direction (v < 0) and has a > 0, its speed |v| decreases (slowing down).
Ex 2.8 Bouncing Ball Speed-Time Plot

2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

First drop: t1 = 4.29 s, v1 = 42 m s−1 • First rebound: v1' = 37.8 m s−1, t2 = 3.86 s (reaches floor at t = 12.0 s)
First Fall (from 90 m):
t1 = √(2h/g) = √(180 / 9.8) = 4.29 s.
Impact speed v1 = g t1 = 9.8 × 4.286 = 42.0 m s−1.
First Rebound:
Rebound speed v1' = 42 − 4.2 = 37.8 m s−1.
Ascent time to peak = v1'/g = 37.8 / 9.8 = 3.86 s (reaches peak at t = 4.29 + 3.86 = 8.15 s).
Descent time = 3.86 s → strikes floor again at t = 8.15 + 3.86 = 12.0 s with speed 37.8 m s−1.
Ex 2.9 Distance vs Displacement Inequality

2.9 Explain clearly, with examples, the distinction between:
(a) Magnitude of displacement and total path length.
(b) Magnitude of average velocity and average speed.
Show that path length ≥ |displacement| and average speed ≥ |average velocity|. When is equality true?

(a) Displacement vs. Path length:
Displacement is the straight-line vector from initial to final position. Path length is the actual distance covered along the trajectory. Since the straight line is the shortest distance between two points, Path Length ≥ |Displacement|.
(b) Average speed vs. |Average velocity|:
Average speed = Path Length / Δt ≥ |Displacement| / Δt = |Average Velocity|.
Condition for Equality:
Equality holds if and only if the particle moves along a straight line in a single unchanging direction without turning back.
Ex 2.10 Average Speed vs Velocity Calculation

2.10 A man walks on a straight road from home to a market 2.5 km away with a speed of 5 km h−1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h−1. What is the magnitude of average velocity and average speed over:
(i) 0 to 30 min  |  (ii) 0 to 50 min  |  (iii) 0 to 40 min?

(i) 0 to 30 min: Avg Vel = 5 km/h, Avg Speed = 5 km/h • (ii) 0 to 50 min: Avg Vel = 0 km/h, Avg Speed = 6 km/h • (iii) 0 to 40 min: Avg Vel = 1.875 km/h, Avg Speed = 5.625 km/h
(i) 0 to 30 min (0.5 h):
Reaches market at 2.5 km. Displacement = 2.5 km, Path length = 2.5 km.
|Avg Velocity| = 2.5 / 0.5 = 5 km h−1. Avg Speed = 5 km h−1.
(ii) 0 to 50 min (5/6 h):
Returns home. Displacement = 0. Path length = 2.5 + 2.5 = 5.0 km.
|Avg Velocity| = 0 km h−1. Avg Speed = 5.0 / (5/6) = 6.0 km h−1.
(iii) 0 to 40 min (2/3 h):
In remaining 10 min (1/6 h), walks back: 7.5 × (1/6) = 1.25 km.
Position from home = 2.5 − 1.25 = 1.25 km. Total path length = 2.5 + 1.25 = 3.75 km.
|Avg Velocity| = 1.25 / (2/3) = 1.875 km h−1.
Avg Speed = 3.75 / (2/3) = 5.625 km h−1.
Ex 2.11 Instantaneous Limit Equivalence

2.11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. Why?

Explanation: Instantaneous speed is limΔt → 0 (Δs/Δt) and magnitude of instantaneous velocity is |limΔt → 0 (Δx/Δt)|. In the infinitesimal limit as Δt → 0, the path length Δs becomes identically equal to the magnitude of the displacement chord |Δx|. Hence, instantaneous speed is always equal to the magnitude of instantaneous velocity.
Ex 2.12 Physically Impossible Graphs

2.12 Look at graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle:

(a) Circle in x-t (b) Loop in v-t (c) Speed < 0 (d) Decreasing Path
Figure 2.10: Graphs (a), (b), (c), and (d) are all physically impossible in one-dimensional motion.
Conclusion: None of the four graphs can represent 1D motion.
(a) Impossible: A vertical line intersects the circle at two points, implying two different positions at the same instant of time.
(b) Impossible: Multi-valued velocity (two different velocities at the same time).
(c) Impossible: Speed is a scalar magnitude and can never be negative.
(d) Impossible: Total path length is monotonically increasing and can never decrease with time.
Ex 2.13 Physical Context of x-t Plot

2.13 Figure 2.11 shows the x-t plot of 1D motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context.

t x O x = 0 (t < 0) x = ½gt² (t > 0)
Figure 2.11: x-t plot representing 1D motion of a particle (rest for t < 0, free fall for t > 0).
Answer: No. The x-t graph represents position along a straight line vs. time, NOT the spatial 2D trajectory of the particle. The motion is strictly 1D along the x-axis.
Physical Context: A ball dropped freely from rest from the top of a tower at t = 0 (stationary x = 0 for t < 0, falling under constant gravity with x = ½gt2 for t > 0).
Ex 2.14 Police & Thief Bullet Speed

2.14 A police van moving on a highway with a speed of 30 km h−1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h−1. If the muzzle speed of the bullet is 150 m s−1, with what speed does the bullet hit the thief’s car?

Answer: Impact speed = 105 m s−1
Step 1: Convert speeds to m s−1.
vvan = 30 × (5/18) = 25/3 m s−1 ≈ 8.33 m s−1.
vthief = 192 × (5/18) = 160/3 m s−1 ≈ 53.33 m s−1.
Step 2: Bullet speed with respect to ground.
vbullet,ground = vmuzzle + vvan = 150 + 25/3 = 475/3 m s−1.
Step 3: Relative speed hitting thief's car.
vrel = vbullet,ground − vthief = (475/3) − (160/3) = 315/3 = 105 m s−1.
Ex 2.15 Physical Scenarios for Graphs

2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
(a) An x-t graph where position stays positive, slope alternates sign rapidly, and amplitude decays.
(b) A v-t graph with sawtooth periodic sign reversals.
(c) An a-t graph with a single brief impulse spike.

(a) Damped Bounces (x-t) (b) Elastic Bouncing (v-t) (c) Impact Pulse (a-t)
Figure 2.12: Physical motion graphs: (a) x-t bouncing ball, (b) v-t elastic rebounds, (c) a-t brief impact pulse.
(a) Physical situation: A ball bouncing repeatedly on the floor, losing energy at each rebound until coming to rest.
(b) Physical situation: A ball dropped on the floor rebounding elastically with reversed velocity sign at each impact.
(c) Physical situation: A cricket ball hit hard by a bat, experiencing a large impulsive acceleration for a very short collision duration.
Ex 2.16 SHM Signs of Variables

2.16 Figure 2.13 gives the x-t plot of a particle executing 1D simple harmonic motion. Give the signs of position (x), velocity (v), and acceleration (a) at t = 0.3 s, 1.2 s, −1.2 s.

t (s) x (m) 0 t = −1.2 s t = 0.3 s t = 1.2 s
Figure 2.13: x-t plot for 1D simple harmonic motion with points indicated at t = 0.3 s, 1.2 s, and −1.2 s.
Time t Position (x) Velocity (v = slope) Acceleration (a = −ω2x)
t = 0.3 s< 0 (−)< 0 (−, negative slope)> 0 (+)
t = 1.2 s> 0 (+)> 0 (+, positive slope)< 0 (−)
t = −1.2 s< 0 (−)> 0 (+, positive slope)> 0 (+)
Ex 2.17 Interval Comparison on x-t Graph

2.17 Figure 2.14 gives the x-t plot of a particle in 1D motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

t x Interval 1 (+) Interval 2 (−) Interval 3 (Fastest)
Figure 2.14: x-t curve divided into 3 equal time intervals for comparing average speed and velocity signs.
Speed: Greatest in Interval 3, Least in Interval 2 • Avg Velocity signs: Interval 1 (+), Interval 2 (−), Interval 3 (+)
Average Speed (Magnitude of slope):
Slope magnitude is steepest in Interval 3 (Greatest speed) and flattest in Interval 2 (Least speed).
Sign of Average Velocity (Direction of slope):
• Interval 1: Slope > 0 → Positive (+).
• Interval 2: Slope < 0 → Negative (−).
• Interval 3: Slope > 0 → Positive (+).
Ex 2.18 Speed-Time Analysis & Vertex Acceleration

2.18 Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which is average speed greatest? Give signs of v and a. What are accelerations at points A, B, C, D?

t Speed A B C D Interval 1 Interval 2 (Max |a|) Interval 3 (Max Speed)
Figure 2.15: Speed-time graph with marked intervals 1, 2, 3 and stationary vertex points A, B, C, D where acceleration a = 0.
Acceleration magnitude greatest in Interval 2 • Average speed greatest in Interval 3 • Accelerations at A, B, C, D are all 0 m s−2
Greatest |Acceleration|: The slope is steepest in Interval 2.
Greatest Average Speed: Height of the curve is maximum in Interval 3.
Signs in Intervals (motion along constant + direction):
• Interval 1: v > 0 (+), a > 0 (+, upward slope).
• Interval 2: v > 0 (+), a < 0 (−, downward slope).
• Interval 3: v > 0 (+), a ≈ 0.
Accelerations at Points A, B, C, D:
Points A, B, C, D are turning peaks/troughs where the tangent is horizontal (slope = 0). Thus, aA = aB = aC = aD = 0.
04 / Rapid Reference

Chapter Summary & Formulas

Comprehensive kinematic rules, calculus shortcuts, graph interpretation cheatsheets, stopping distances, relative motion in 1D, and exam tips.

100% SYLLABUS
2.1 & 2.2 • Position, Distance, Displacement

Position, Displacement & Speed vs Velocity

Motion in a straight line is 1D motion. It requires a frame of reference (origin and coordinate axis).

  • Path Length (Distance): Total length of path traversed by an object. It is a scalar, always positive ($\ge 0$), and never decreases with time.
  • Displacement ($\Delta x$): Change in position: $\Delta x = x_f - x_i$. It is a vector (indicated by $+$ or $-$ in 1D), and can be positive, negative, or zero.
Inequality: $\text{Path Length} \ge |\text{Displacement}|$. They are equal only if the object moves in a straight line without reversing direction.

Velocity and Speed

Quantity Mathematical Formula Crucial Conditions & Interpretation
Average Velocity $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}$ Depends only on initial and final positions. Independent of path. Can be zero if starting and ending points are same.
Average Speed $s_{\text{avg}} = \frac{\text{Total Path Length}}{\text{Total Time}}$ Always greater than or equal to the magnitude of average velocity: $s_{\text{avg}} \ge |v_{\text{avg}}|$.
Instantaneous Velocity $v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$ Defined at a specific instant. Geometrically, it is the slope of the tangent to the position-time ($x-t$) graph at that instant.
Instantaneous Speed $s = |v| = \left|\frac{dx}{dt}\right|$ Always equal to the magnitude of instantaneous velocity. Measured by a vehicle's speedometer.
2.3 • Acceleration

Acceleration & Calculus in Kinematics

Acceleration is the rate of change of velocity with respect to time.

  • Average Acceleration: $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$
  • Instantaneous Acceleration: $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
The Space Derivative Trick: By Chain Rule: $$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \implies a = v\frac{dv}{dx}$$ Use this formula when acceleration or velocity is given as a function of position $x$ instead of time $t$.

Speeding Up vs. Slowing Down Condition

The sign of acceleration alone does NOT tell you whether speed is increasing or decreasing:
  • Speeding Up (Acceleration): Velocity and acceleration have the same sign (both $+$, or both $-$).
  • Slowing Down (Deceleration): Velocity and acceleration have opposite signs (one $+$, one $-$).
  • Example: An object moving in negative direction ($v < 0$) with negative acceleration ($a < 0$) is actually speeding up!
Kinematic Graphs • Geometrical Rules

Kinematic Graph Interpretation Cheat Sheet

Graph Type Slope Meaning Area Meaning
Position-time ($x-t$) Slope $= \text{Velocity } (v)$ No physical meaning
Velocity-time ($v-t$) Slope $= \text{Acceleration } (a)$
  • Algebraic Area $= \text{Displacement } (\Delta x)$
  • Total Magnitude Area $= \text{Distance}$
Acceleration-time ($a-t$) No physical meaning (slope is jerk) Area $= \text{Change in velocity } (v_f - v_i)$
Curvature Rule ($x-t$ Graph): Concave Upward ($\smile$, positive curvature) $\implies a > 0$.
Concave Downward ($\frown$, negative curvature) $\implies a < 0$.
2.4 • Kinematic Equations

Equations of Motion for Uniform Acceleration

Strict Condition: These equations are valid only when acceleration $a$ is constant throughout the motion.

Equation of Motion Variables Included
$$v = v_0 + at$$ $v, v_0, a, t$ (No displacement $x$)
$$x = v_0 t + \frac{1}{2}at^2$$ $x, v_0, a, t$ (No final velocity $v$)
$$v^2 = v_0^2 + 2ax$$ $v, v_0, a, x$ (No time $t$)
$$D_n = v_0 + \frac{a}{2}(2n - 1)$$ Distance traveled strictly in the $n^{\text{th}}$ second.

Vertical Motion Under Gravity (Free Fall)

Taking upward direction as positive ($+$) and downward as negative ($-$), we have $a = -g \approx -9.8\text{ m/s}^2$:

  • Time of Ascent ($t_a$) / Descent ($t_d$): $t_a = t_d = \frac{u}{g}$
  • Total Time of Flight ($T$): $T = \frac{2u}{g}$
  • Maximum Height Reached ($H_{\text{max}}$): $H_{\text{max}} = \frac{u^2}{2g}$
Galileo's Law of Odd Numbers: The distances traversed by a freely falling body (starting from rest) in equal, successive intervals of time $\tau$ bear to one another the ratio of the odd numbers starting from unity: $$\text{Ratio} = 1 : 3 : 5 : 7 : 9 : \dots$$
2.4 & 2.5 • Applications

Stopping Distance, Reaction Time & Relative Velocity

Stopping Distance ($d_s$)

The distance a vehicle travels before coming to rest when braking acceleration (deceleration $-a$) is applied:

$$d_s = \frac{v_0^2}{2a}$$
Trick: Stopping distance is proportional to the square of initial velocity ($d_s \propto v_0^2$). If you double your speed, the stopping distance increases by a factor of four!

Reaction Time ($t_r$)

The time a person takes to observe, think, and initiate action. If a ruler is dropped and falls a distance $d$ before being caught: $$t_r = \sqrt{\frac{2d}{g}}$$

Relative Velocity in 1D

The velocity of object $B$ with respect to object $A$ is: $$v_{BA} = v_B - v_A$$

  • Objects moving in the same direction: Subtract their magnitudes ($v_{BA} = v_B - v_A$).
  • Objects moving in opposite directions: Add their magnitudes ($v_{BA} = v_B + v_A$).
NCERT Official

Points to Ponder & Exam Tips

  1. The origin and positive direction of an axis are choices. Specify this choice before assigning signs to displacement, velocity, and acceleration.
  2. At the peak of a vertical throw, velocity is zero but acceleration is NOT zero ($a = -g$).
  3. Calculus definitions ($v = dx/dt$ and $a = dv/dt$) are exact and always true. Standard equations of motion are ONLY valid for constant acceleration.
  4. The relative velocity of $A$ w.r.t $B$ is the negative of the relative velocity of $B$ w.r.t $A$ ($v_{AB} = -v_{BA}$).
  5. In relative motion, acceleration of B relative to A is $a_{BA} = a_B - a_A$.
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to calculate score and review answers.

24 QUESTIONS
1. The slope of a position-time graph gives:
2. The area under a velocity-time graph represents:
3. The SI unit of acceleration is:
4. At the highest point of a vertically projected ball, velocity is:
5. If speed is constant in 1D motion, acceleration is:
6. In Galileo's law of odd numbers, the ratio of successive interval distances is:
7. The relative velocity formula for B with respect to A is:
8. For uniform acceleration, the position-time graph is a:
Active Test Level