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

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.

Worked example

1. 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.
2. Note that the change is not 2 × (5 − 3) = 4; ignoring the reversal of direction is the standard error.

First law (law of inertia)

• 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.
• 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.
• Inertia is the property by which 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.

Third law

• 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.

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.

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.

Instantaneous velocity

• v = lim_Δt→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 from the velocity curve

• 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.

Worked example

1. 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.

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.

Acceleration as a function of position

• 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).

Accelerated and decelerated motion

• 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.

Worked examples

1. 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.
2. 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.
3. Same idea with a square root: if v = √(2s + 1), then ½v² = s + ½ and a = d/ds(½v²) = 1 m/s², the same at every position.

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.

Worked example

1. 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.
2. Path elimination: 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.

From acceleration to velocity to position

• Integration undoes differentiation: v = ∫ a dt and s = ∫ v dt.
• Between two instants: v(t₂) − v(t₁) = ∫_t₁^t₂ a dt, and displacement = ∫_t₁^t₂ 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.

Worked example

1. 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.

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.

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.

Units of force

• 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.

Worked examples

1. 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.
2. 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.

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.

Worked examples

1. 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.
2. 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.
3. 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.

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.

Worked examples

1. A force F = 6t newtons acts on a body along a line. Δp = ∫₀⁴ 6t dt = [3t²]₀⁴ = 48 kg·m/s.
2. 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.

Several forces

• 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.

1. s = t³ − 3t² + 4 (m). v at t = 2 = 0 m/s; a at t = 2 = 6 m/s²
2. v = 2s² + 3. a = v dv/ds = (2s² + 3)(4s), so at s = 1: a = 20 m/s²
3. 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
4. r = (2t² − t) i + (t³) j. v at t = 1 = 3 i + 3 j, |v| = 3√2
5. 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
6. 3 kg-wt expressed in newtons = 29.4 N; in dynes = 2.94 × 10⁶ dyne
7. Uniform motion in a straight line implies the resultant force = zero
8. Distance from t = 0 to t = 3 when v = t² − 4t + 3: 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