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Overview

Express the position, displacement, velocity and acceleration of a particle as functions of time — or of position — and pass from one to another by differentiation and integration.
Analyse motion along a straight line graphically from the (position–time) and (velocity–time) curves, and obtain the distance travelled.
Use momentum, Newton’s three laws, the units of force, and the equations of motion for bodies of constant and of variable mass.

Vector Functions, Calculus and Newton's Laws
├── Momentum and its change
├── Newton's first and third laws
├── Position vector, displacement and distance
├── Average and instantaneous velocity
├── Acceleration (average, instantaneous, as a function of position)
├── Vector functions in the plane → differentiate component by component
├── Integration to recover velocity and position
├── Graphical analysis of motion
├── Newton's second law and the equation of motion
│   ├── Constant mass: F = m a
│   └── Units of force: newton, dyne, kg-wt, g-wt
├── Variable-mass motion: F = m dv/dt + v dm/dt
└── Integration with the second law: Δp = ∫ F dt

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 1 / 14

Momentum and its change

Momentum

The momentum of a body is the product of its mass and its velocity vector: p = m v; it is a vector quantity pointing in the direction of the velocity.
Its unit is kg·m/s in the absolute system, and g·cm/s in the centimetric (C.G.S.) system.
The change of momentum is the final momentum minus the initial one: Δp = m v₂ − m v₁.
In motion along a straight line choose one direction as positive; velocities in the opposite direction enter the calculation with a negative sign.

A body of mass 2 kg moves at 3 m/s to the right, strikes a wall and returns at 5 m/s to the left. Taking right as positive: v₁ = +3, v₂ = −5, so Δp = 2(−5) − 2(3) = −10 − 6 = −16 kg·m/s, i.e. a change of 16 kg·m/s directed to the left.
Note that the change is not 2 × (5 − 3) = 4; ignoring the reversal of direction is the standard error.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 2 / 14

Newton’s first and third laws

First law (law of inertia)

Comparison aspect	Main idea	Keywords
•	A body remains at rest or continues to move with uniform velocity in a straight line unless a resultant force acts on it to change that state	bodyremainsrestcontinues
Converse use	if a body moves with uniform velocity then the resultant force on it is zero — this is how resistance to motion is found from the engine force	bodymovesuniformvelocity
Inertia	is the property by a body resists any change in its state of motion and it increases with mass. A train reaches its maximum speed when the engine force equals the total resistance so a = 0	propertybodyresistschange

For every action there is a reaction equal in magnitude and opposite in direction.
Action and reaction act on two different bodies, which is why they never cancel each other.
Example the Earth’s reaction to a body resting on it equals the pressure the body exerts on the Earth.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 3 / 14

Position vector, displacement and distance

Motion along a straight line

We describe motion of a particle on a line with all quantities written in terms of the time t.
The position vector r(t) locates the particle from the origin; its magnitude is measured in units of length. We use s (or x) for the algebraic measure of the position, so its sign gives the side of the origin.
The displacement vector is the change of position between two instants: s = r(t₂) − r(t₁).
Distance is the actual length of the path travelled; it is scalar and is never negative, whereas displacement may be positive, negative or zero.
A body that returns to its starting point has zero displacement but non-zero distance.

Differentiation and Integration of Vector: a ball flying on a curved path shows vector functions and Newtons laws of motion: r(1) = (10, 5) and maximum height 5 m.

Point of comparison	Displacement	Distance
Type of quantity	Vector	Scalar
Sign	positive, negative or zero	always positive (or zero)
Depends on	only the initial and final points	the whole path

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 4 / 14

Average and instantaneous velocity

Average velocity

v_av = displacement ÷ elapsed time = [s(t₂) − s(t₁)]/(t₂ − t₁).
Geometrically it is the slope of the chord joining the two points of the (position–time) curve.

v = limΔt→0 [s(t + Δt) − s(t)]/Δt = ds/dt velocity is the first derivative of position, i.e. the slope of the tangent to the (position–time) curve.
The (scalar) speed is the magnitude |v| and is never negative. v > 0 means motion in the positive direction, v < 0 the opposite.
The velocity vanishes at the instant the particle reverses its direction of motion.
Average speed = total distance ÷ total time, and it is generally different from the magnitude of the average velocity.

Distance in [t₁, t₂] is the area between the (velocity–time) curve and the time axis: d = ∫_t₁^t₂ |v| dt.
When the sign of v changes, split the interval at the instants where v = 0 and add the absolute values.

A stone is thrown vertically upward so that its height after t seconds is s = 49t − 4.9t². Find the greatest height and the velocity when the height is 78.4 m.
v = ds/dt = 49 − 9.8t. At the top v = 0 → t = 5 s, s_max = 245 − 122.5 = 122.5 m.
49t − 4.9t² = 78.4 → divide by 4.9: t² − 10t + 16 = 0 → (t − 2)(t − 8) = 0 → t = 2 or 8.
v(2) = 49 − 19.6 = +29.4 m/s (rising), v(8) = 49 − 78.4 = −29.4 m/s (falling); the speed is 29.4 m/s in both cases.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 5 / 14

Acceleration

Average and instantaneous

a_av = Δv/Δt; the instantaneous acceleration is a = dv/dt = d²s/dt², the slope of the tangent to the (velocity–time) curve. Unit: m/s² (or cm/s²).
From s(t) differentiate once for v, twice for a.

When v is given in terms of s, use the chain rule→a = dv/dt = (dv/ds)(ds/dt) = v dv/ds, equivalently a = d/ds (½v²).

Motion is accelerated when |v| increases — that is when v and a have the same sign.
Motion is decelerated (retarded) when |v| decreases — when v and a have opposite signs.
a may be zero at an instant while the body is still moving that happens at a maximum or a minimum velocity.

A particle moves on a line with s = t³ − 6t² + 9t (metres). v = 3t² − 12t + 9 = 3(t − 1)(t − 3), a = 6t − 12.
(a) v = 0 at t = 1 and t = 3: a(1) = −6 m/s², a(3) = +6 m/s².
(b) a = 0 at t = 2: v = 12 − 24 + 9 = −3 m/s, so the speed is 3 m/s.
(c) Direction changes at t = 1 and t = 3 with s(0) = 0, s(1) = 4, s(3) = 0, s(4) = 4, so distance = 4 + 4 + 4 = 12 m.
A particle moves with v = 2s + 1. Find a when s = 3 m: a = v dv/ds = (2s + 1)(2), so at s = 3, a = 7 × 2 = 14 m/s².
if v = √(2s + 1), then ½v² = s + ½ and a = d/ds(½v²) = 1 m/s², the same at every position.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 6 / 14

Vector functions in the plane

Differentiation component by component

If the position vector is r(t) = x(t) i + y(t) j, its derivative is found by differentiating each component: v = dr/dt = ẋ i + ẏ j and a = dv/dt = ẍ i + ÿ j.
The magnitude of velocity is |v| = √(ẋ² + ẏ²); the magnitude of acceleration is found the same way.
Integration is also done component by component, and each component needs its own constant — together they form a constant vector.

A particle moves so that r = (t³ − 3t) i + 2t² j. Then v = (3t² − 3) i + 4t j and a = 6t i + 4 j.
At t = 2: v = 9 i + 8 j and a = 12 i + 4 j.
|v| = √(81 + 64) = √145 ≈ 12.04 m/s.
from x = t³ − 3t and y = 2t² the trajectory can be studied by eliminating t; differentiating is usually far quicker than finding the cartesian path.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 7 / 14

Integration to find velocity and position

From acceleration to velocity to position

Integration undoes differentiation v = ∫ a dt and s = ∫ v dt.
Between two instants v(t₂) − v(t₁) = ∫t₁t₂ a dt, and displacement = ∫t₁t₂ v dt.
The two constants of integration come from the initial conditions (initial velocity and initial position at t = 0).
When a is constant these relations reduce to the familiar equations of motion: v = v₀ + at, s = v₀t + ½at², v² = v₀² + 2as.

A particle has a = 6t − 4 m/s², initial velocity 2 m/s and initial position 1 m.
v = ∫(6t − 4)dt = 3t² − 4t + C; at t = 0, v = 2 → C = 2, so v = 3t² − 4t + 2.
s = ∫v dt = t³ − 2t² + 2t + C′; at t = 0, s = 1 → C′ = 1, so s = t³ − 2t² + 2t + 1.
At t = 2: v = 12 − 8 + 2 = 6 m/s and s = 8 − 8 + 4 + 1 = 5 m.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 8 / 14

Graphical analysis of motion

Reading the curves

(position–time) curve the slope of the tangent gives the velocity; a point with a horizontal tangent is an instant of zero velocity.
(velocity–time) curve the slope of the tangent gives the acceleration, and the area under the curve gives the displacement.
If the velocity graph is a straight (inclined) line the acceleration is constant; if it is a horizontal line the velocity is uniform and the acceleration is zero.
Area above the time axis is positive displacement, area below it is negative displacement; to get distance add the areas with their absolute values.
The intersection of two (velocity–time) curves marks the instant when the two bodies have equal velocities — not necessarily when they meet.

Curve	Slope gives	Area gives
s–t	velocity	—
v–t	acceleration	displacement (signed), \|area\| sum = distance
a–t	rate of change of acceleration	change in velocity

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 9 / 14

Newton’s second law and the equation of motion

Statement

The rate of change of momentum of a body is proportional to the resultant force acting on it and takes place in the direction of that force. With suitable units the constant of proportionality is 1: F = d(mv)/dt.
If the mass is constant, the equation of motion becomes F = m dv/dt = m a: the resultant force produces an acceleration in its own direction, directly proportional to the force and inversely proportional to the mass.
Mass is a constant scalar measuring inertia; weight is the force of the Earth’s attraction and changes with position: W = m g, with g ≈ 9.8 m/s².

Unit	Definition	Value
newton (N)	force giving 1 kg an acceleration of 1 m/s²	1 kg·m/s²
dyne	force giving 1 g an acceleration of 1 cm/s²	1 g·cm/s²
kg-wt (kgf)	weight of a mass of 1 kg	9.8 N
g-wt (gf)	weight of a mass of 1 g	980 dyne

Between the two absolute units 1 N = 10⁵ dyne.
In F = ma only absolute units are used, so a force given in kg-wt or g-wt must first be converted.

A constant force of 12 N acts on a 3 kg body at rest. a = 12/3 = 4 m/s²; v = v₀ + at = 0 + 4(5) = 20 m/s after 5 s.
A force of 2 kg-wt acts on a 5 kg body. F = 2 × 9.8 = 19.6 N, so a = 19.6/5 = 3.92 m/s².

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 10 / 14

Motion of bodies of variable mass

The general equation

When the mass changes with time (a body collecting dust or burning fuel) we must not write F = m a; the product must be differentiated: F = d(mv)/dt = m dv/dt + v dm/dt.
The term v dm/dt appears only because the mass changes; it vanishes for constant mass and we return to F = m a.
Procedure write m as a function of t, get v by differentiating s, form p = m v, then differentiate p to get F.

A body of mass (2t + 1) kg moves on a line with s = t² metres. v = ds/dt = 2t m/s, so p = m v = (2t + 1)(2t) = 4t² + 2t and F = dp/dt = 8t + 2.
At t = 3: p = 36 + 6 = 42 kg·m/s and F = 24 + 2 = 26 N.
A metal ball of mass 100 g moves with uniform velocity 10 m/s through dust that sticks to it at 2 g/s. Then m = 100 + 2t grams. Uniform velocity means a = 0, so F = v dm/dt. With v = 10 m/s = 1000 cm/s and dm/dt = 2 g/s: F = 1000 × 2 = 2000 dyne, constant and in the direction of motion.
Reading the phrase “collects dust at a constant rate” as dm/dt = 0 is wrong: it is precisely what makes dm/dt a positive constant.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 11 / 14

Using integration with the second law

From force or acceleration to the change of momentum

Because F = dp/dt, the change of momentum between two instants is the integral of the force over time: Δp = p₂ − p₁ = ∫_t₁^t₂ F dt.
For a body of constant mass Δp = m∫a dt = m Δv.
Use this whenever the force or the acceleration is given as a function of time.

A force F = 6t newtons acts on a body along a line. Δp = ∫₀⁴ 6t dt = [3t²]₀⁴ = 48 kg·m/s.
A 5 kg body moves with acceleration a = 4t m/s². F = ma = 20t N, so at t = 2, F = 40 N; and Δp = ∫₀³ 20t dt = [10t²]₀³ = 90 kg·m/s.

When several forces act, the resultant R is their vector sum, then R = m a.
In vector form the law is applied to each component separately: R_x = m ẍ, R_y = m ÿ.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 12 / 14

Test yourself

s = t³ − 3t² + 4 (m). v at t = 2 = 0 m/s; a at t = 2 = 6 m/s²
v = 2s² + 3. a = v dv/ds = (2s² + 3)(4s), so at s = 1: a = 20 m/s²
m = 3t + 2 kg, s = t²/2 m. v = t, so p = m v = 3t² + 2t; at t = 2, p = 16 kg·m/s and F = dp/dt = 6t + 2 = 14 N
r = (2t² − t) i + (t³) j. v at t = 1 = 3 i + 3 j, |v| = 3√2
A 4 kg body has v = 6 m/s and a force F = 2t N acts for 5 s. Δp = 25 kg·m/s, final velocity = 12.25 m/s
3 kg-wt expressed in newtons = 29.4 N; in dynes = 2.94 × 10⁶ dyne
Uniform motion in a straight line implies the resultant force = zero
v vanishes at t = 1 and t = 3; ∫v = t³/3 − 2t² + 3t gives +4/3 on [0, 1] and −4/3 on [1, 3], so the distance is 8/3 m while the displacement is 0

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 13 / 14

Exam traps and key takeaways

💡 High-yield traps

Momentum is a vector assign a positive direction first, otherwise a rebound gives Δp = m(v₂ − v₁) with the wrong sign.
Distance is the integral of |v|, not of v; split the interval where v changes sign.
Zero velocity at an instant does not mean zero acceleration (a ball at its highest point still has a = 9.8 m/s²).
Accelerated versus decelerated is decided by the signs of v and a, not by the sign of a alone.
For variable mass, F = m dv/dt + v dm/dt; using F = ma alone is marked wrong.
Convert kg-wt and g-wt to newtons or dynes before F = ma, and cm/s² with grams gives dynes, not newtons (1 N = 10⁵ dyne).
Weight changes with position, mass never does — a common “give a reason” question.
On a (v–t) graph, slope = acceleration and area = displacement; on an (s–t) graph, slope = velocity. Mixing the two axes loses the whole part.
When integrating twice, introduce two constants and use both initial conditions; losing the constant of position is a frequent half-mark.
Action and reaction never cancel because they act on different bodies.

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion · 14 / 14

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Differentiation and Integration of Vector Functions and Newton’s Laws of Motion