Saturday, August 15, 2026

Kinematics Chapter 2 – Complete Definitions, Formulas, Derivations and Graphs

Kinematics Chapter 2: Master Revision Guide

Definitions · formulas · derivations · graphs · worked examples · common mistakes

How to use this page: Read the quick formula sheet first. For revision, try to recall each definition or derivation before opening its detailed explanation.

1. Symbols and SI Units

SymbolMeaningSI unitType
ddistancemetre (m)scalar
sdisplacementmetre (m)vector
uinitial velocitym/svector
vfinal velocitym/svector
aaccelerationm/s²vector
ttimesecond (s)scalar
gmagnitude of gravitational acceleration9.8 m/s² (approximately 10)downward

2. Quick Formula Sheet

Speed
speed = distance/time
Velocity
velocity = displacement/time
Acceleration
a = (v − u)/t
First equation
v = u + at
Second equation
s = ut + ½at²
Third equation
v² = u² + 2as
Average velocity
v̄ = total displacement/total time
nth-second distance
Sₙ = u + (a/2)(2n − 1)
Vertical projection
t↑ = u/g; H = u²/(2g); T = 2u/g
Dropped body
v = gt; h = ½gt²; v² = 2gh

3. Definitions

Motion, rest and reference point

Motion: a body is in motion when its position changes with time relative to a chosen reference point. Rest: its position does not change relative to that reference. Motion is relative: a seated passenger is at rest relative to the bus but moving relative to the road.

Scalar and vector quantities

A scalar has magnitude only: distance, speed, time and mass. A vector has magnitude and direction: displacement, velocity and acceleration. “5 m” is a scalar distance; “5 m east” is a vector displacement.

Distance and displacement

Distance is the total path length travelled. It is scalar and never negative. Displacement is the straight-line change from initial to final position, with direction. Its magnitude is always ≤ distance.

Example: walk 3 m east and 3 m west. Distance = 6 m; displacement = 0.
Speed, velocity and their averages

Speed is distance covered per unit time. Velocity is displacement per unit time in a specified direction. Average speed = total distance/total time. Average velocity = total displacement/total time. Instantaneous speed or velocity is the value at one particular instant.

Uniform and non-uniform motion

Uniform velocity: equal displacements in equal time intervals in the same direction. Non-uniform velocity: unequal displacements in equal times, or a changing direction. Uniform speed in a circle is not uniform velocity because direction continuously changes.

Acceleration and retardation

Acceleration is rate of change of velocity: a = Δv/Δt. It occurs when speed changes, direction changes, or both. Retardation/deceleration means acceleration opposite to the velocity, causing speed to decrease. Negative acceleration does not automatically mean slowing; the relative directions matter.

Types of motion
  • Translational: the whole body changes position; rectilinear along a straight line or curvilinear along a curve.
  • Rotational: body turns about an axis, such as a wheel.
  • Circular: object follows a circular path.
  • Oscillatory: repeated to-and-fro motion about a mean position, such as a pendulum.
  • Random: irregular path, such as a dust particle in air.
Projectile, trajectory and free fall

A projectile is an object that, after being projected, moves under gravity alone (air resistance ignored). Its path is the trajectory. An oblique projectile is launched at an angle to the horizontal. Free fall is motion under gravity alone. Near Earth, all freely falling bodies have the same acceleration g if air resistance is neglected.

4. Equations of Uniformly Accelerated Motion

Condition: These equations apply only when acceleration is constant and motion is along one straight line. Choose a positive direction first and keep the signs consistent.
Derivation 1: v = u + at

Acceleration is change in velocity per unit time:

a = (v − u)/t

at = v − u

v = u + at

Example: u = 4 m/s, a = 3 m/s², t = 5 s. Then v = 4 + 3(5) = 19 m/s.
Derivation 2: s = ut + ½at² (average-velocity method)

For constant acceleration, average velocity = (u + v)/2.

s = [(u + v)/2]t

Using v = u + at:

s = [(u + u + at)/2]t = [u + ½at]t

s = ut + ½at²

Example: u = 5 m/s, a = 2 m/s², t = 4 s. s = 5(4) + ½(2)(4²) = 36 m.
Graphical derivation: s = ut + ½at²
tvutrectangle = uttriangle = ½(v − u)t

Displacement equals area under the velocity-time graph:

s = rectangle + triangle = ut + ½(v − u)t. Since v − u = at, s = ut + ½at².

Derivation 3: v² = u² + 2as

Start with s = [(u + v)/2]t and use t = (v − u)/a:

s = [(u + v)/2][(v − u)/a]

2as = (u + v)(v − u) = v² − u²

v² = u² + 2as

Example: u = 3 m/s, a = 2 m/s², s = 10 m. v² = 9 + 40 = 49, so v = 7 m/s.
Distance travelled in the nth second

Total displacement after n seconds: sₙ = un + ½an². After n − 1 seconds: sₙ₋₁ = u(n − 1) + ½a(n − 1)².

Sₙ = sₙ − sₙ₋₁

Sₙ = u + (a/2)(2n − 1)

Example: u = 3 m/s, a = 2 m/s², n = 4. S₄ = 3 + (2/2)(7) = 10 m.

5. Motion Graphs

Displacement–timeslope = velocitytimes Velocity–timeslope = accelerationarea = displacementtimev Acceleration–timearea = change in vtimea At rest: s–thorizontal: position fixed Constant velocity: v–thorizontal: a = 0
  • s–t graph: slope gives velocity; horizontal means rest; straight slope means uniform velocity; curved line means changing velocity.
  • v–t graph: slope gives acceleration; signed area gives displacement; horizontal means constant velocity.
  • a–t graph: signed area gives change in velocity; line on the time axis means zero acceleration.

6. Gravity and Vertical Motion

Sign convention for g

The sign depends on the chosen positive direction, not on whether the object is moving upward or downward.

  • If upward is positive: a = −g throughout.
  • If downward is positive: a = +g throughout.

At maximum height v = 0, but acceleration remains g downward.

Body dropped from rest

Put u = 0 and a = g in the motion equations:

v = gt; h = ½gt²; v² = 2gh.

Example: dropped for 3 s with g = 9.8 m/s². v = 29.4 m/s downward and h = 44.1 m.
Vertical projection: time of ascent, maximum height and total time

Choose upward positive, so a = −g. At the top, v = 0.

From v = u − gt: t↑ = u/g.

From v² = u² − 2gH: H = u²/(2g).

If it returns to the same level, descent time equals ascent time: T = 2u/g.

Eliminating u using u = gT/2 gives H = gT²/8.

Example: u = 19.6 m/s. t↑ = 2 s, H = 19.6 m and total time T = 4 s.

7. Average Speed and Velocity

Equal time intervals

If velocities v₁ and v₂ act for equal times, average velocity = (v₁ + v₂)/2.

10 m/s for 4 s and 18 m/s for 4 s gives average = 14 m/s.
Equal distances

For equal distances travelled at speeds v₁ and v₂:

Let each distance be d. Total distance = 2d; total time = d/v₁ + d/v₂.

average speed = 2v₁v₂/(v₁ + v₂)

Equal distances at 40 and 60 km/h: average speed = 2(40)(60)/(100) = 48 km/h. If the trip returns to its start, average velocity = 0.
Three equal distances

For equal distances at speeds v₁, v₂ and v₃:

average speed = 3/(1/v₁ + 1/v₂ + 1/v₃).

At 10, 20 and 30 km/h: average = 3/(1/10 + 1/20 + 1/30) = 180/11 ≈ 16.36 km/h.

8. Projectile Motion Essentials

uu cos θu sin θmaximum heightrange R

Ignoring air resistance, horizontal velocity remains constant while vertical velocity changes due to gravity. Components: uₓ = u cosθ, uᵧ = u sinθ. For launch and landing at the same level:

  • time of flight: T = 2u sinθ/g
  • maximum height: H = u²sin²θ/(2g)
  • range: R = u²sin2θ/g

Range is maximum at 45° because sin90° = 1. Complementary angles such as 30° and 60° have equal range because sin60° = sin120°, although 60° goes higher and stays airborne longer.

9. Common Mistakes Checklist

  • Do not confuse distance with displacement or speed with velocity.
  • Convert units before substituting: 1 km/h = 5/18 m/s.
  • Choose a positive direction and keep signs consistent.
  • At maximum height, velocity is zero but acceleration is not.
  • The standard three equations require constant acceleration.
  • Area below a velocity-time axis is negative displacement.
  • For average speed, use total distance/total time—not usually the simple arithmetic mean.
  • In circular motion, constant speed can still mean changing velocity.

10. Mixed Worked Revision Example

Problem: A car begins at 5 m/s and accelerates uniformly at 2 m/s² for 6 s. Find final velocity, displacement and distance in the sixth second.

Final velocity: v = u + at = 5 + 2(6) = 17 m/s.
Displacement: s = ut + ½at² = 5(6) + ½(2)(36) = 66 m.
Sixth-second distance: S₆ = u + (a/2)(2×6 − 1) = 5 + 1(11) = 16 m.
Check: distance in first 5 s = 5(5) + ½(2)(25) = 50 m; 66 − 50 = 16 m.
Final memory map: slope of s–t → velocity; slope of v–t → acceleration; area under v–t → displacement; area under a–t → change in velocity.

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