Chapter 2: Motion in a Straight Line | Class 11 Physics | AAI ATC Exam Prep

📌 Chapter Overview

📍 Position, Displacement & Distance

⚡ Instantaneous Velocity & Speed

📈 Acceleration & v-t Graphs

🔢 3 Kinematic Equations

🍎 Free Fall & Projectile (1D)

↔️ Relative Velocity

Position, Displacement & Average Velocity

Kinematics studies how objects move — without asking why. An object in motion is treated as a point object when its size is much smaller than the distances involved.

Key Definitions

Quantity	Definition	Formula	SI Unit
Position (x)	Location with respect to origin	—	m
Distance	Total path length (scalar)	Path length	m
Displacement (Δx)	Change in position (vector)	x₂ − x₁	m
Average speed	Total path / total time (scalar)	d/Δt	m/s
Average velocity	Displacement / time interval	Δx/Δt	m/s

      📌 Average speed ≥ |Average velocity| always. Equality holds only when path is straight with no backtracking.

      📌 Position right of origin → positive; left → negative (by convention).

🎯 A car travels 5 km east then 3 km west: Distance = 8 km, but Displacement = 2 km east. These are never interchangeable!

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ADD IMAGE HERE
x-t graph showing position vs time — slope = velocity (Fig. 2.1)

🎯 Practice MCQs — Position & Average Velocity

Q1 A woman walks 2.5 km to her office at 5 km/h and returns by auto at 25 km/h. Her average speed for the entire journey is:

A 15 km/h

B 8.33 km/h

C 12.5 km/h

D 0 km/h

✅ Time to office = 2.5/5 = 0.5 h; Time to return = 2.5/25 = 0.1 h; Total time = 0.6 h
Total distance = 5 km; Average speed = 5/0.6 = 8.33 km/h
Note: Average velocity = 0 (displacement = 0, returns home)

Q2 Position of an object: x = a + bt² where a = 8.5 m, b = 2.5 m/s². Average velocity between t = 2.0 s and t = 4.0 s is:

A 5 m/s

B 10 m/s

C 15 m/s

D 20 m/s

✅ x(2) = 8.5 + 2.5(4) = 18.5 m; x(4) = 8.5 + 2.5(16) = 48.5 m
Avg velocity = (48.5 − 18.5)/(4 − 2) = 30/2 = 15 m/s

Q3 A man walks 2.5 km to a market at 5 km/h. The market is closed, so he returns home at 7.5 km/h. His average speed over the entire 0 to 50 min interval is:

A 0 km/h

B 6.25 km/h

C 6 km/h

D 5 km/h

✅ Time to market = 2.5/5 = 0.5 h = 30 min. Remaining time = 50−30 = 20 min = 1/3 h
Distance returned = 7.5 × (1/3) = 2.5 km
Total path = 2.5 + 2.5 = 5 km; Time = 50/60 h
Average speed = 5/(5/6) = 6 km/h

Instantaneous Velocity & Speed

Instantaneous velocity is the velocity at a specific instant — the limiting value of average velocity as Δt → 0.

v = lim(Δt→0) Δx/Δt = dx/dt Instantaneous speed = |v| = magnitude of instantaneous velocity

      📈 On a position-time (x-t) graph:

      — Instantaneous velocity at any instant = slope of the tangent to the curve at that point

      — For uniform motion: x-t graph is a straight line; slope = constant velocity

      — For accelerated motion: x-t graph is a parabola (curves upward or downward)

Graph Reading Tip: Steeper slope → greater speed. Negative slope → moving in negative direction. Zero slope (horizontal line) → object at rest.

⚠️ Average speed over a finite interval is always ≥ |average velocity|. But instantaneous speed = |instantaneous velocity| always — they are equal at every instant.

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ADD IMAGE HERE
x-t graph showing tangent at t=4s gives instantaneous velocity (Fig. 2.1 & Table 2.1)

🎯 Practice MCQs — Instantaneous Velocity

Q4 For x = 0.08t³ m, the instantaneous velocity at t = 4 s is:

A 3.2 m/s

B 3.92 m/s

C 3.84 m/s

D 0.08 m/s

✅ v = dx/dt = d(0.08t³)/dt = 0.24t²
At t = 4 s: v = 0.24 × 16 = 3.84 m/s
(This matches the limiting value from Table 2.1 in NCERT)

Q5 For x = a + bt² where b = 2.5 m/s², the instantaneous velocity at t = 2.0 s is:

A 2.5 m/s

B 5.0 m/s

C 10.0 m/s

D 15.0 m/s

✅ v = dx/dt = 2bt = 2 × 2.5 × t = 5.0t m/s
At t = 2.0 s: v = 5.0 × 2 = 10.0 m/s

Q6 On a position-time graph, if the slope of the tangent at a point is negative, it means:

A The object is at rest

B The object is decelerating

C The object is moving in the negative direction

D The acceleration is negative

✅ Slope of x-t graph = instantaneous velocity. A negative slope means the position is decreasing with time → object is moving in the negative (opposite) direction.

Acceleration & Velocity-Time Graphs

Acceleration is the rate of change of velocity with time. It was Galileo who first established that free fall has constant acceleration — not constant velocity with distance.

Average acceleration: ā = Δv/Δt = (v₂ − v₁)/(t₂ − t₁) [m/s²] Instantaneous acceleration: a = lim(Δt→0) Δv/Δt = dv/dt

v-t Graph Interpretations

      📊 On a velocity-time (v-t) graph:

      — Slope at any point = instantaneous acceleration

      — Area under the v-t curve = displacement over that time interval

      — Horizontal line → uniform motion (a = 0)

      — Straight inclined line → uniform acceleration

      — Curved v-t graph → non-uniform acceleration

Types of Motion — x-t Graph Shapes

Motion Type	x-t graph	v-t graph	a-t graph
Uniform (a=0)	Straight line (inclined)	Horizontal line	Horizontal at 0
Positive acceleration	Parabola (curves up)	Straight line (rising)	Horizontal (positive)
Negative acceleration	Parabola (curves down)	Straight line (falling)	Horizontal (negative)

🎯 Zero velocity at an instant does NOT mean zero acceleration. A ball thrown up has v = 0 at the highest point, but acceleration = g = 9.8 m/s² downward throughout!

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ADD IMAGE HERE
v-t graphs for 4 types of accelerated motion showing area = displacement (Fig. 2.3, 2.4)

🎯 Practice MCQs — Acceleration & v-t Graphs

Q7 A car starts from rest and reaches 35 m/s in 7 s with uniform acceleration. Its acceleration is:

A 7 m/s²

B 5 m/s²

C 35 m/s²

D 245 m/s²

✅ a = Δv/Δt = (35 − 0)/7 = 5 m/s²

Q8 A v-t graph shows a straight line from v = 20 m/s at t = 0 to v = 0 at t = 4 s. The displacement during this interval equals (area under graph):

A 80 m

B 40 m

C 20 m

D 10 m

✅ Area under v-t graph = area of triangle = ½ × base × height = ½ × 4 × 20 = 40 m

Q9 A car moving at 126 km/h is stopped within 200 m. The retardation (uniform) is:

A 5 m/s²

B 3.0625 m/s²

C 35 m/s²

D 6.125 m/s²

✅ v₀ = 126 km/h = 35 m/s; v = 0; x = 200 m
v² = v₀² + 2ax → 0 = 35² + 2a(200) → a = −1225/400 = −3.0625 m/s²
Retardation = 3.0625 m/s². Time to stop = (0−35)/(−3.0625) ≈ 11.4 s

Kinematic Equations for Uniform Acceleration

For motion with constant acceleration, three kinematic equations relate the five quantities: displacement (x), time (t), initial velocity (v₀), final velocity (v), and acceleration (a).

Equation 1: v = v₀ + at Equation 2: x = v₀t + ½at² Equation 3: v² = v₀² + 2ax General form (starting from x₀): x = x₀ + v₀t + ½at² v² = v₀² + 2a(x − x₀)

      💡 These equations are derived from integration of dv/dt = a and dx/dt = v.

      Average velocity for uniform acceleration: v̄ = (v₀ + v)/2

      Stopping distance: dₛ = v₀²/(2a) — proportional to square of initial velocity!

      Doubling speed → 4× stopping distance — critical for road safety.

Galileo's Law of Odd Numbers

For a body falling from rest, distances covered in successive equal time intervals (τ) are in the ratio 1 : 3 : 5 : 7 : 9 : 11... (odd numbers).

y = −½gτ², −4/2gτ², −9/2gτ², −16/2gτ², ... Successive distances: y₀, 3y₀, 5y₀, 7y₀, ... (ratio 1:3:5:7...)

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ADD IMAGE HERE
v-t graph showing derivation of x = v₀t + ½at² as area under curve (Fig. 2.5)

🎯 Practice MCQs — Kinematic Equations

Q10 A ball is thrown upward at 20 m/s from a 25 m tall building. How high above the throw point does it rise? (g = 10 m/s²)

A 25 m

B 20 m

C 45 m

D 40 m

✅ At maximum height, v = 0; v₀ = 20 m/s; a = −10 m/s²
v² = v₀² + 2a(y − y₀) → 0 = 400 + 2(−10)(y − y₀)
(y − y₀) = 400/20 = 20 m

Q11 The same ball (thrown from 25 m building at 20 m/s upward, g = 10 m/s²) hits the ground after:

A 2 s

B 3 s

C 5 s

D 4 s

✅ Taking upward as +y, origin at ground: y₀ = 25 m, v₀ = +20 m/s, a = −10 m/s², y = 0
0 = 25 + 20t − 5t² → 5t² − 20t − 25 = 0 → t² − 4t − 5 = 0
(t − 5)(t + 1) = 0 → t = 5 s (taking positive root)

Q12 A player throws a ball at 29.4 m/s upward. To what maximum height does it rise? (g = 9.8 m/s²)

A 29.4 m

B 44.1 m

C 58.8 m

D 88.2 m

✅ v² = v₀² − 2gH → 0 = (29.4)² − 2(9.8)H
H = 864.36/(2 × 9.8) = 864.36/19.6 = 44.1 m

Free Fall & Applications

Free fall = motion under gravity alone (no air resistance). It is a special case of uniformly accelerated motion with a = g = 9.8 m/s² (downward).

Taking downward as negative (upward +y), released from rest (v₀ = 0): v = −gt = −9.8t m/s y = −½gt² = −4.9t² m v² = −2gy = −19.6y m²/s² Free fall acceleration: g = 9.8 m/s² ≈ 10 m/s² (approx)

Key Applications

      🚗 Stopping distance: dₛ = v₀²/(2a) — braking distance quadruples when speed doubles

      ⏱️ Reaction time measurement: Ruler drop experiment — tᵣ = √(2d/g)

      If d = 21 cm: tᵣ = √(2 × 0.21/9.8) ≈ 0.2 s

🎯 At the highest point of projection: velocity = 0 but acceleration = g (downward). Zero velocity ≠ zero acceleration! This is a very common exam trap.

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ADD IMAGE HERE
Free fall graphs — a-t, v-t, and y-t plots showing constant −9.8 m/s² (Fig. 2.7)

🎯 Practice MCQs — Free Fall

Q13 An object is dropped from rest at height 80 m. How long does it take to reach the ground? (g = 10 m/s²)

A 2 s

B 3 s

C 4 s

D 8 s

✅ h = ½gt² → 80 = ½(10)t² → t² = 16 → t = 4 s

Q14 In the ruler-drop reaction time experiment, a ruler falls 21.0 cm before being caught. The reaction time is: (g = 9.8 m/s²)

A 0.1 s

B ≈ 0.2 s

C 0.5 s

D 1.0 s

✅ d = ½gtᵣ² → tᵣ = √(2d/g) = √(2 × 0.21/9.8) = √(0.042/9.8) = √(0.0429) ≈ 0.207 s ≈ 0.2 s

Q15 Galileo's law of odd numbers: a body falls from rest in successive equal time intervals τ. The distances traversed in intervals 1, 2, 3 are in ratio:

A 1 : 2 : 3

B 1 : 4 : 9

C 1 : 3 : 5

D 2 : 4 : 6

✅ y₁ = ½gτ², y₂ = ½g(2τ)² − ½gτ² = 3 × ½gτ², y₃ = 5 × ½gτ²
Ratio of distances in successive intervals = 1 : 3 : 5 (odd numbers)

Relative Velocity

The velocity of one object as observed from another moving object is called relative velocity. It depends on the reference frame of the observer.

Velocity of A relative to B: v_AB = v_A − v_B If both moving in SAME direction: |v_AB| = |v_A − v_B| (smaller) If moving in OPPOSITE directions: |v_AB| = |v_A + v_B| (larger)

Practical Example — Bullet & Thief's Car

Police van: 30 km/h = 30/3.6 = 8.33 m/s; Bullet muzzle speed: 150 m/s; Thief's car: 192 km/h = 53.33 m/s

Bullet speed (ground frame) = police speed + muzzle speed = 8.33 + 150 = 158.33 m/s Speed of bullet relative to thief's car: = 158.33 − 53.33 = 105 m/s

      ✈️ AAI ATC Relevance: Relative velocity is fundamental to aviation — pilots must account for wind speed (relative to ground), airspeed (relative to air), and groundspeed when navigating. ATC uses ground speed for separation standards.
    

🌟 Far-away objects like Moon and stars seem stationary from a fast train because their angular velocity relative to the observer is negligible — a real-world consequence of relative velocity and large distances.

🎯 Practice MCQs — Relative Velocity

Q16 A police van at 30 km/h fires a bullet (muzzle speed 150 m/s) at a thief's car moving at 192 km/h in the same direction. The speed of the bullet relative to the thief's car is:

A 150 m/s

B 96.67 m/s

C 105 m/s

D 203.33 m/s

✅ Police van = 30/3.6 = 8.33 m/s; Thief's car = 192/3.6 = 53.33 m/s
Bullet speed (ground) = 8.33 + 150 = 158.33 m/s
Relative to thief = 158.33 − 53.33 = 105 m/s

Q17 Train A moves east at 60 m/s; Train B moves west at 40 m/s. Speed of B relative to A is:

A 20 m/s

B 60 m/s

C 100 m/s

D 40 m/s

✅ Taking east as +ve: v_A = +60 m/s; v_B = −40 m/s
v_BA = v_B − v_A = −40 − 60 = −100 m/s
Speed = |−100| = 100 m/s (opposite directions add up)

Q18 A drunkard takes 5 steps forward and 3 steps backward repeatedly. Each step = 1 m, takes 1 s. How long does it take to fall into a pit 13 m away?

A 13 s

B 20 s

C 32 s

D 37 s

✅ Each cycle (5 forward + 3 backward) = 8 s, net progress = 2 m
After 4 cycles (32 s): position = 8 m
At t=32s: starts 5th forward journey from 8 m
Steps: 9m(33s), 10m(34s), 11m(35s), 12m(36s), 13m(37s) → falls in pit at t = 37 s

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🏃 Motion in a Straight Line