LINEAR MOTION 1/3 free

CHAPTER 1: LINEAR MOTION

Linear motion is the most fundamental type of motion—movement along a straight line. Everything from a car speeding down a highway to an apple falling from a tree obeys the laws described in this chapter. Master this, and you master the language of mechanics.

1.1 DISPLACEMENT, VELOCITY, AND ACCELERATION

Before we can analyze motion, we must define the tools we use. These are not just words; they are precise, measurable quantities.

s
Displacement
Straight-line distance from start to finish, with direction. Vector.
v
Velocity
Rate of change of displacement (m/s). Vector.
a
Acceleration
Rate of change of velocity (m/s²). Vector.
Start A End B Distance (actual path length) Displacement (straight line from start to finish)
Distance versus displacement: The curved path shows total distance travelled; the straight arrow shows displacement.

1.1.1 Displacement (s)

  • Definition: Displacement is the distance moved in a specified direction. It is the straight-line distance from an object's starting point to its ending point.
  • Scalar vs. Vector: Displacement is a vector (magnitude + direction). Distance is a scalar (only magnitude).
  • SI Unit: metre (m)
  • Example: Walking 5 km North → displacement = 5 km North. Walk a circle back to start → displacement = ZERO.
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Critical Point: Always distinguish between distance (how much ground covered) and displacement (how far and direction from start). Displacement can be zero even when distance is large.

1.1.2 Velocity (v)

  • Definition: Velocity is the rate of change of displacement.
  • Vector quantity; speed is scalar.
  • Average velocity: v = Δs / Δt
  • Example: 20 m/s East ≠ 20 m/s West — velocities differ.

1.1.3 Acceleration (a)

  • Definition: Rate of change of velocity.
  • Vector — same direction as change in velocity.
  • Average acceleration: a = Δv / Δt
  • Positive acceleration = speeding up in positive direction; negative = deceleration; zero = constant velocity.
t=0 v=0 t=1 v=10 t=2 v=20 t=3 v=30 d₁ d₂ > d₁ d₃ > d₂ Increasing displacement each second → positive acceleration
Positive acceleration: The car covers more distance in each successive time interval.
Worked Example 1: Acceleration from Rest
Problem:
A car starts from rest and reaches 20 m/s East in 5 s. Find acceleration.
Given:
u = 0, v = 20 m/s, t = 5 s
Formula:
a = (v-u)/t
Answer:
a = (20-0)/5 = 4 m/s² East
Worked Example 2: Deceleration (Braking)
Problem:
Train at 30 m/s stops in 10 s. Find acceleration.
Given:
u = 30 m/s, v = 0, t = 10 s
Solution:
a = (0-30)/10 = -3 m/s² (deceleration)

1.2 TICKER-TIMER EXPERIMENTS

Before high-speed cameras, scientists used ticker-timers — a classic experiment linking theory to measurement.

TAPE 30 mm 30 mm 30 mm TROLLEY Ticker timer makes dots at regular time intervals (0.02 s) Equal spacing = constant velocity Increasing spacing = acceleration
Ticker timer apparatus: The tape passes through the timer which marks dots at fixed time intervals. Dot spacing reveals motion.

1.2.1 What is a Ticker-Timer?

A ticker-timer makes dots on paper tape at regular intervals (usually 0.02 s). Attached to a moving object, the dot spacing reveals motion.

  • Frequency: 50 Hz → 50 dots/sec
  • Period (T): T = 1/f = 0.02 s

1.2.2 Analyzing the Tape

  • Constant Velocity: Evenly spaced dots → same speed each interval
  • Acceleration: Increasing gaps → speeding up
  • Deceleration: Decreasing gaps → slowing down
Constant v: Acceleration: Deceleration: Equal spacing → constant velocity Increasing spacing → acceleration
Three ticker tape patterns: constant velocity (top), acceleration (middle), deceleration (bottom).
Worked Example 3: Ticker Tape Analysis
Given:
f=50 Hz, 10 spaces: first length=0.2 m, later length=0.5 m, time between centers=0.4 s
Time for 10 spaces:
10 × 0.02 = 0.2 s
Initial velocity u:
0.2 m / 0.2 s = 1 m/s
Final velocity v:
0.5 m / 0.2 s = 2.5 m/s
Acceleration:
a = (2.5 - 1)/0.4 = 3.75 m/s²

1.3 MOTION GRAPHS

Time (s) Displacement (m) s = ut (constant velocity) increasing slope t₁ t₂ t₃
Displacement-Time graph: Straight line = constant velocity (slope = v). Curved line = acceleration (changing slope).

1.3.1 Displacement-Time (s-t) Graphs

  • Gradient = velocity — steeper slope means higher speed
  • Straight line → constant velocity
  • Curved line → changing velocity (acceleration or deceleration)
  • Horizontal line → zero velocity (object at rest)
Time (s) Velocity (m/s) a = slope a = 0 Area = displacement
Velocity-Time graph: Gradient = acceleration. Area under graph = displacement travelled.

1.3.2 Velocity-Time (v-t) Graphs

  • Gradient = acceleration — slope indicates rate of change of velocity
  • Area under graph = displacement — total distance covered
  • Horizontal line → constant velocity (zero acceleration)
  • Positive slope → acceleration; negative slope → deceleration
Worked Example 4: Graph Interpretation
Problem:
v-t graph: (0,0) to (5,20) then horizontal to (10,20). Describe motion.
Answer:
0–5 s: constant acceleration a = 4 m/s²; 5–10 s: constant velocity 20 m/s

1.4 EQUATIONS OF LINEAR MOTION (THE "BIG THREE")

For constant acceleration only.

EquationContainsMissing variable
v = u + atu, v, a, ts (displacement)
s = ut + ½at²s, u, a, tv (final velocity)
v² = u² + 2asv, u, a, st (time)
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CRITICAL RULE: These equations ONLY work for motion with constant (uniform) acceleration.
Ground t = 0 s v = 0 m/s t = 1 s, v = 10 m/s ↓ t = 2 s, v = 20 m/s ↓ v increases a = g = 9.8 m/s² downward
Free fall under gravity: All objects accelerate downward at g = 9.8 m/s², regardless of mass (ignoring air resistance).

1.5 MOTION UNDER GRAVITY: FREE FALL

Near Earth's surface, all objects fall with constant acceleration g ≈ 9.8 m/s² (≈10 m/s²) directed downward.

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Pro Tip: Choose a sign convention before solving and stick with it. Downwards = positive OR upwards = positive.
Worked Example 8: Object Dropped
Height:
s = ½ × 10 × 9 = 45 m
Worked Example 9: Thrown Upwards
Max height:
s = 20²/(2×10) = 20 m
Total time:
t = 2u/g = 4 s

✍️ EXTENSIVE PRACTICE QUESTIONS

1
Distinguish between distance and displacement.
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Answer
Distance: scalar (magnitude only). Displacement: vector (magnitude + direction).
2
A car travels 100 m East, then 40 m West. Find distance and displacement.
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Answer
Distance = 140 m, Displacement = 60 m East.
3
A stone is dropped from 45 m. Time to hit ground? (g=10 m/s²)
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Answer
t = √(2s/g) = 3 s
4
Car accelerates from rest at 2 m/s² for 8 s. Find final velocity and distance.
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Answer
v = 16 m/s, s = 64 m.
5
Ball thrown upward at 15 m/s. Find maximum height. (g=10 m/s²)
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Answer
s = 11.25 m.

Selected Answers

1. Scalar vs vector  |  2. 140 m, 60 m East  |  3. 3 s
4. 16 m/s, 64 m  |  5. 11.25 m
g ≈ 9.8 or 10 m/s²  |  At highest point: v=0, a = g downward

Linear motion is the gateway to all of physics. Master this chapter, and you master the language of mechanics.

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MAGNETISM THE GAS LAWS