Further Kinematics

M2 TOPIC MAP

What You Need to Master

Further Kinematics in Mechanics 2 extends the M1 treatment of motion. The key change is that you are no longer restricted to constant acceleration. You must be comfortable moving between displacement, velocity and acceleration using calculus.

01 — Kinematics foundations 02 — Variable acceleration 03 — x(t), v(t), a(t) 04 — Integration & initial conditions 05 — Vertical motion 06 — Projectiles 07 — Vector kinematics 08 — 100-question exam

LESSON 01

Kinematics: The Language of Motion

Kinematics describes the motion of a particle without asking what forces caused that motion. In Mechanics 2, the particle may move in a straight line or in a plane.

Scalar quantities

A scalar has magnitude only.

Distance
Speed
Time
Mass

Vector quantities

A vector has magnitude and direction.

Displacement
Velocity
Acceleration
Position vector

Particle model

A particle is treated as having mass but negligible size. Its shape and rotational behaviour are ignored. This lets us concentrate on translational motion.

Core quantities

Quantity	Symbol	SI unit	Meaning
Position	x or r	m	Location relative to an origin
Displacement	s or Δx	m	Change in position
Velocity	v	m s⁻¹	Rate of change of displacement
Acceleration	a	m s⁻²	Rate of change of velocity

v = dx/dt Velocity is the derivative of displacement with respect to time.

a = dv/dt = d²x/dt² Acceleration is the derivative of velocity with respect to time.

Exam habit:

Always identify what the question gives you first: x(t), v(t), a(t), an initial condition, a position, a velocity, or a time interval. Then decide whether differentiation or integration is required.

LESSON 02

Variable Acceleration

In M1, the standard constant-acceleration formulae are often enough. In M2, acceleration may depend on time, so its value changes throughout the motion.

Constant acceleration

The acceleration has the same value at every instant.

Variable acceleration

The acceleration changes with time or, when represented componentwise, changes according to the motion model.

The fundamental chain

x(t) → differentiate → v(t) → differentiate → a(t)

a(t) → integrate → v(t) → integrate → x(t)

Do not lose the constants.

Every indefinite integration introduces a constant of integration. The initial conditions determine those constants.

Worked Example — Variable Acceleration

A particle has acceleration

a = 6t + 4

Given that its velocity is 3 m s⁻¹ when t = 0, find v in terms of t.

Since a = dv/dt,

v = ∫(6t + 4) dt

v = 3t² + 4t + C

At t = 0, v = 3.

Therefore C = 3.

v = 3t² + 4t + 3

LESSON 03

When Displacement Is Given as a Function of Time

One of the most direct M2 questions gives the displacement of a particle as a function of time.

x = f(t)

Differentiate once to obtain velocity and twice to obtain acceleration.

v = dx/dt

a = d²x/dt²

Worked Example

A particle moves so that

x = 2t³ - 9t² + 12t + 5

Find its velocity and acceleration.

Differentiate:

v = 6t² - 18t + 12

Differentiate again:

a = 12t - 18

Finding when the particle is stationary

A particle is instantaneously at rest when its velocity is zero.

v = 0 Use this equation to find stationary instants.

Finding position at a given time

If x(t) is known, simply substitute the required value of t. If only v(t) is known, integrate and use an initial position.

Common trap:

“Stationary” does not mean “zero acceleration”. A particle can have v = 0 while a ≠ 0.

LESSON 04

Integration and Initial Conditions

Integration reverses differentiation. If acceleration is known, integrate to obtain velocity. If velocity is known, integrate again to obtain displacement.

v = ∫a dt

x = ∫v dt

Why initial conditions matter

Suppose

a = 12t²

Then

v = 4t³ + C

Without additional information, C cannot be determined.

Worked Example — Two Integrations

A particle has acceleration

a = 12t² - 6t

At t = 0, the particle has velocity 5 m s⁻¹ and displacement 2 m. Find x in terms of t.

First integrate:

v = 4t³ - 3t² + C

Since v(0) = 5,

C = 5

Therefore

v = 4t³ - 3t² + 5

Integrate again:

x = t⁴ - t³ + 5t + D

Since x(0) = 2,

D = 2

x = t⁴ - t³ + 5t + 2

Fast method:

Write the initial condition immediately beside the equation. This reduces the chance of forgetting which constant it determines.

LESSON 05

Displacement, Distance and Direction

In one-dimensional motion, displacement can be positive or negative. Distance travelled is never negative.

Displacement

Net change in position:

Δx = x₂ - x₁

Distance

Total length of the path travelled.

It requires care if the particle changes direction.

Detecting a change of direction

In one-dimensional motion, direction changes when velocity changes sign. Therefore solve v = 0 and check the sign of v on either side.

Worked Example — Distance Travelled

A particle has

v = t² - 4t + 3

Find the times at which the particle is stationary.

Set v = 0:

t² - 4t + 3 = 0

(t - 1)(t - 3) = 0

Therefore

t = 1 and t = 3

These times divide the motion into intervals whose directions must be checked separately when finding total distance.

Important:

If a question asks for total distance, do not simply calculate final displacement when the particle reverses direction.

LESSON 06

Vertical Motion Under Gravity

Vertical motion is one of the most important applications of constant acceleration. When air resistance is ignored, the acceleration due to gravity is constant.

g = 9.8 m s⁻² Unless the question explicitly states another value.

Choose a positive direction

Upward positive

Velocity upwards is positive.

a = -9.8

Downward positive

Velocity downwards is positive.

a = +9.8

Either convention is valid. The danger is changing conventions halfway through a calculation.

Vertical motion equations

v = u + at

s = ut + ½at²

v² = u² + 2as

Worked Example — Vertical Projection

A particle is projected vertically upwards with speed 19.6 m s⁻¹. Find its maximum height above its point of projection.

Take upwards as positive.

At maximum height, v = 0.

u = 19.6, a = -9.8.

Use:

v² = u² + 2as

0 = 19.6² - 19.6s

s = 19.6 m

Maximum height condition:

At the highest point, the instantaneous velocity is zero. The acceleration is still -9.8 m s⁻².

LESSON 07

Projectile Motion

A projectile is a particle projected into the air and then allowed to move under gravity, with air resistance neglected.

The central idea:

Horizontal and vertical motion are treated separately. They are connected only because both use the same time t.

Resolving the initial velocity

If a particle is projected with speed u at an angle θ above the horizontal:

Horizontal component = u cosθ

Vertical component = u sinθ

Horizontal motion

If horizontal acceleration is zero:

x = u cosθ · t

vₓ = u cosθ

Vertical motion

Taking upward as positive:

y = u sinθ · t − ½gt²

vᵧ = u sinθ − gt

Trajectory equation

Eliminating t gives the equation of the projectile path.

y = x tanθ − gx² / (2u²cos²θ)

Same-level projectile formulae

For a projectile launched and landing at the same vertical level:

Time of flight = 2u sinθ / g

Range = u² sin(2θ) / g

Maximum height = u²sin²θ / (2g)

Conditions matter:

The standard range and total-time formulae above assume the launch and landing points are at the same height.

Worked Example — Projectile Components

A particle is projected at 20 m s⁻¹ at 30° above the horizontal. Find its initial horizontal and vertical velocity components.

Horizontal:

20cos30° = 17.32 m s⁻¹

Vertical:

20sin30° = 10.00 m s⁻¹

Initial velocity = 17.32i + 10j m s⁻¹

LESSON 08

Projectile Motion: Full Exam Method

Choose axes.

Usually take horizontal as x and vertical as y.

Resolve the initial velocity.

Write u cosθ horizontally and u sinθ vertically.

Write separate equations.

Horizontal and vertical motion have different equations.

Use the same time.

The particle reaches a particular point at one value of t, regardless of which component equation you use.

Substitute the geometric information.

Height, range or horizontal distance gives the required equation for t or another unknown.

Check the physical answer.

Time should normally be positive, and the sign of each velocity component should match the direction of motion.

Worked Example — Horizontal Distance

A particle is projected at 14 m s⁻¹ at 45° above the horizontal. Find its horizontal displacement after 1 second.

Horizontal velocity:

vₓ = 14cos45°

vₓ ≈ 9.899 m s⁻¹

Horizontal acceleration is zero, so:

x = vₓt

x ≈ 9.90 m

LESSON 09

Vector Kinematics

M2 extends kinematics into two-dimensional vector form. Instead of describing position with one coordinate x, we can describe it with a position vector.

r = xi + yj

Velocity vector

v = dr/dt

Therefore, if

r = x(t)i + y(t)j

then

v = x′(t)i + y′(t)j

Acceleration vector

a = dv/dt = d²r/dt²

Worked Example — Vector Differentiation

A particle has position vector

r = (t² + 3t)i + (2t³ − t)j

Find its velocity and acceleration vectors.

Differentiate once:

v = (2t + 3)i + (6t² − 1)j

Differentiate again:

a = 2i + 12tj

Speed

Speed is the magnitude of velocity.

|v| = √(vₓ² + vᵧ²)

Acceleration magnitude

|a| = √(aₓ² + aᵧ²)

Position at a particular time

Substitute the time directly into r(t). The resulting vector gives the particle's position relative to the chosen origin.

Vector rule:

Differentiate or integrate each component independently. Never differentiate only the scalar magnitude when the question asks for the vector itself.

LESSON 10

Integrating a Vector

Integration works component-by-component exactly as differentiation does.

If a = f(t)i + g(t)j, then v = ∫f(t)dt i + ∫g(t)dt j

Worked Example — From Acceleration to Position

A particle has acceleration

a = 6ti + 4j

Its initial velocity is

v(0) = 2i − 3j

Find v(t).

Integrate both components:

v = (3t² + A)i + (4t + B)j

At t = 0:

2i − 3j = Ai + Bj

Therefore A = 2 and B = −3.

v = (3t² + 2)i + (4t − 3)j

Component trap:

A constant of integration is required for each independent component unless the initial condition determines it directly.

LESSON 11

Vector Displacement Between Two Times

If the position vector is known, the displacement from t = t₁ to t = t₂ is simply the difference between the two position vectors.

Displacement = r(t₂) − r(t₁)

Worked Example

A particle has

r = t²i + (t³ − 2t)j

Find its displacement from t = 1 to t = 3.

At t = 3:

r(3) = 9i + 21j

At t = 1:

r(1) = i − j

Therefore:

r(3) − r(1) = 8i + 22j

Displacement = 8i + 22j m

LESSON 12

Speed, Direction and Stationary Motion in Vector Form

In two-dimensional motion, a particle is stationary when its entire velocity vector is zero.

v = 0 ⇔ vₓ = 0 and vᵧ = 0

Speed is:

speed = |v| = √(vₓ² + vᵧ²)

Direction of velocity

If the velocity vector is

v = vₓi + vᵧj

the angle α measured from the positive x-axis satisfies

tanα = vᵧ / vₓ

The quadrant must be considered when interpreting the direction.

Do not use tan⁻¹ blindly.

A calculator may return an angle in the wrong quadrant. Use the signs of vₓ and vᵧ to identify the actual direction.

LESSON 13

Constant Acceleration in Vector Form

When acceleration is constant as a vector, the familiar constant-acceleration equations can be written in vector form.

v = u + at

r = r₀ + ut + ½at²

These equations are especially useful when the initial velocity and acceleration are vectors.

Worked Example

A particle starts at the origin with

u = 4i + 3j

and has constant acceleration

a = 2i − j

Find its position after 2 seconds.

Use:

r = ut + ½at²

r = (4i + 3j)(2) + ½(2i − j)(4)

r = 8i + 6j + 4i − 2j

r = 12i + 4j m

LESSON 14

Projectile Motion as Vector Motion

Projectile motion can be written elegantly using vectors. If a particle is projected from the origin with initial speed u at angle θ, then:

r = (u cosθ)t i + (u sinθ)t j − ½gt² j

Equivalently:

r = (u cosθ)t i + (u sinθ · t − ½gt²)j

Velocity vector

v = u cosθ i + (u sinθ − gt)j

Acceleration vector

a = −gj

Key observation:

Gravity affects only the vertical component in the standard projectile model. Horizontal acceleration is zero.

LESSON 15

Projectile Motion: Maximum Height

At maximum height, the vertical component of velocity is zero. The horizontal component remains unchanged.

vᵧ = u sinθ − gt

Set vᵧ = 0:

t = u sinθ / g

Substitute this time into the vertical displacement equation to obtain the maximum height.

H = u²sin²θ / (2g)

Important:

At maximum height, the particle is not “motionless” in general. Its vertical velocity is zero, but its horizontal velocity remains u cosθ.

LESSON 16

Projectile Motion: Time of Flight and Range

For a projectile returning to its launch height:

T = 2u sinθ / g

Horizontal range is horizontal speed multiplied by time of flight:

R = u cosθ × T

Hence:

R = u²sin2θ / g

Worked Example

A projectile is fired at 20 m s⁻¹ at 30° and lands at the same height. Find its time of flight.

T = 2u sinθ / g

T = 40 × 0.5 / 9.8

T ≈ 2.04 s

LESSON 17

Trajectory Equation

Sometimes the question asks for the equation of the path rather than the position at a particular time.

Start with:

x = u cosθ · t

Rearrange:

t = x/(u cosθ)

Substitute into

y = u sinθ · t − ½gt²

giving:

y = x tanθ − gx²/(2u²cos²θ)

Exam structure:

If asked for the trajectory, eliminate t. If asked for the particle's position at a time, keep t.

LESSON 18

Advanced Exam Method

Translate the words into variables.

Identify x, y, u, v, a and t.

Choose a positive direction.

Especially important for vertical motion.

Differentiate if you have position.

x(t) → v(t) → a(t).

Integrate if you have acceleration or velocity.

Always include constants for indefinite integrals.

Use initial conditions.

Substitute known values of t, x or v.

For projectiles, separate components.

Horizontal and vertical equations share the same time.

For vectors, work component-by-component.

Differentiate or integrate i and j components separately.

Check the physical meaning.

A negative velocity may be perfectly valid; a negative time normally indicates that the mathematical root is not part of the physical interval being considered.

LESSON 19

High-Frequency Exam Traps

Trap 01

Forgetting the constant of integration.

Trap 02

Confusing displacement with distance.

Trap 03

Using v = 0 to claim acceleration is also zero.

Trap 04

Using same-level projectile formulae when launch and landing heights differ.

Trap 05

Forgetting that the horizontal component of projectile velocity is constant.

Trap 06

Differentiating the magnitude instead of the vector components.

LESSON 20

Ultimate Formula Atlas

v = dx/dt

a = dv/dt = d²x/dt²

v = ∫a dt

x = ∫v dt

r = xi + yj

v = dr/dt

a = d²r/dt²

|v| = √(vₓ² + vᵧ²)

|a| = √(aₓ² + aᵧ²)

v = u + at

r = r₀ + ut + ½at²

vᵧ = u sinθ − gt

y = u sinθ · t − ½gt²

x = u cosθ · t

T = 2u sinθ/g

R = u²sin2θ/g

H = u²sin²θ/(2g)

y = x tanθ − gx²/(2u²cos²θ)

IAL ATLAS EXAM LAB