Unit 3: Kinematics

Instantaneous Velocity and Acceleration

In kinematics, motion is described by how an object's position changes with time. While average quantities give an overall picture, instantaneous quantities describe the state of motion at a precise instant.

Instantaneous Velocity

The instantaneous velocity v of a particle is defined as the limit of the average velocity as the time interval approaches zero:

v = lim_{Δt → 0} (Δs / Δt) = ds/dt

Where Δs is the displacement during the small time interval Δt, and ds/dt is the derivative of the position function s(t) with respect to time. Graphically, instantaneous velocity is the slope of the tangent to the s‑t curve at a given point.

Instantaneous Acceleration

Similarly, instantaneous acceleration a is the rate of change of velocity:

a = lim_{Δt → 0} (Δv / Δt) = dv/dt

Here Δv is the change in velocity over Δt. On a v‑t graph, instantaneous acceleration equals the slope of the tangent line.

Distinction from Average Quantities

• Average velocity = total displacement / total time (Δs/Δt) over a finite interval.
• Average acceleration = change in velocity / time interval (Δv/Δt) over a finite interval.
• Instantaneous values converge to the average values only when the motion is uniform (constant velocity or acceleration).

Relative Velocity

When analyzing motion from different reference frames, it is useful to compute the velocity of one object as observed from another.

Definition

The velocity of object A relative to object B is given by vector subtraction:

v_{AB} = v_A - v_B

Where v_A and v_B are the velocities of A and B measured in a common inertial frame.

One‑Dimensional Case

If motion is restricted to a straight line, the relative velocity reduces to simple algebraic subtraction, taking care of sign conventions.

Example: A train moves east at 20 m/s (v_A = +20 m/s) and a car moves west at 15 m/s (v_B = -15 m/s). The velocity of the train relative to the car is v_{AB} = 20 - (-15) = 35 m/s east.

Two‑Dimensional Case

In a plane, velocities are vectors. Subtract them component‑wise:

v_{AB,x} = v_{A,x} - v_{B,x}
v_{AB,y} = v_{A,y} - v_{B,y}

The magnitude and direction follow from Pythagoras and trigonometry.

Applications

1. Rain‑Man Problem: A man walks horizontally with speed v_m while rain falls vertically with speed v_r. The apparent direction of rain relative to the man is given by v_{rain/man} = v_r - v_m. To avoid getting wet, he tilts his umbrella opposite to this relative velocity vector.
2. Boat‑River Problem: A boat aims to cross a river flowing with speed v_r. If the boat’s speed in still water is v_b and it points at an angle θ to the riverbank, the resultant velocity relative to ground is the vector sum v_{boat/ground} = v_b + v_r. Solving for the required heading to reach a point directly opposite uses component equations.

Equations of Motion (Graphical Treatment)

For motion with constant acceleration, three fundamental equations relate displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). They can be derived directly from velocity‑time (v‑t) and displacement‑time (s‑t) graphs.

First Equation: Velocity‑Time Relation

From the definition of constant acceleration:

a = (v - u) / t → v = u + at

Graphically, this is the slope of the v‑t line.

Second Equation: Position‑Time Relation

The displacement equals the area under the v‑t graph. For constant acceleration, the graph is a trapezoid:

s = ut + (1/2) a t^2

Derivation: Area = rectangle (u·t) + triangle (½·a·t·t).

Third Equation: Velocity‑Position Relation

Eliminate t using the first two equations:

v^2 = u^2 + 2as

Graphically, this corresponds to the area of a trapezium expressed in terms of velocities.

Displacement in the nth Second

The distance travelled during the nth second (s_n) is:

s_n = u + a(2n-1)/2

Derived by subtracting the total displacement after (n-1) seconds from that after n seconds.

Summary Table

Motion of a Freely Falling Body

When air resistance is negligible, any object near Earth’s surface experiences a constant downward acceleration due to gravity, denoted g.

Basic Equations (u = 0)

Taking downward as positive and releasing from rest (u = 0):

v = gt
h = ½ gt²
v² = 2gh

Where h is the vertical distance fallen.

Numerical Value of g

Standard average: g ≈ 9.8 m/s². More precisely, g varies with altitude (h) and latitude (φ) according to:

g(φ, h) = g₀ [1 - 2h/R] + 0.0053024 sin²φ - 0.0000058 sin²2φ

where g₀ = 9.80665 m/s² and R is Earth’s radius (~6371 km).

Example Problem

A ball is dropped from a height of 45 m. Find the time to hit the ground and its speed just before impact.

Using h = ½ gt² → t = √(2h/g) = √(2·45/9.8) ≈ 3.03 s.
Speed: v = gt = 9.8·3.03 ≈ 29.7 m/s downward.

Projectile Motion

A projectile is any object launched into the air and subject only to gravity (neglecting air resistance). Its motion can be analyzed as two independent one‑dimensional motions: horizontal (uniform) and vertical (uniformly accelerated).

Horizontal Projection (θ = 0°)

If launched horizontally with speed v₀ from height H:

x = v₀ t (horizontal)
y = ½ gt² (vertical, measured downward from launch point)

Eliminating t gives the trajectory:

y = (g / 2v₀²) x² — a parabola opening downward.

General Launch at Angle θ

Let initial speed be v₀ and launch angle above the horizontal be θ. Then:

uₓ = v₀ cosθ (horizontal component)
uᵧ = v₀ sinθ (vertical component)

Equations of motion:

x = uₓ t = v₀ cosθ · t
y = uᵧ t – ½ gt² = v₀ sinθ · t – ½ gt²

Key Characteristics

• Time of Flight (T): Time until the projectile returns to the same vertical level (y = 0). Solving 0 = v₀ sinθ·T – ½ gT² gives:

T = 2 v₀ sinθ / g

• Horizontal Range (R): Horizontal distance travelled in time T:

R = uₓ T = (v₀ cosθ)·(2 v₀ sinθ / g) = v₀² sin2θ / g

Maximum range occurs when sin2θ = 1 → θ = 45°.

• Maximum Height (H): Peak vertical displacement (vᵧ = 0) reached at time t = v₀ sinθ / g:

H = (v₀ sinθ)² / (2g) = v₀² sin²θ / (2g)

Trajectory Equation

Eliminate t from the parametric equations:

y = x tanθ – (g x²) / (2 v₀² cos²θ)

This is a quadratic in x, confirming a parabolic path.

Applications

1. Sports: Calculating the optimal angle for a basketball free throw, or the speed needed for a football to clear a goalpost.
2. Military: Determining launch angles for artillery shells to hit distant targets.
3. Space Launch: Initial rocket trajectory approximates projectile motion before thrust dominates; understanding gravity turn is essential.

Example Problem

A cricket ball is thrown with an initial speed of 25 m/s at an angle of 30° above the horizontal. Find (a) time of flight, (b) range, and (c) maximum height.

Given: v₀ = 25 m/s, θ = 30°, g = 9.8 m/s².
(a) T = 2·25·sin30° / 9.8 = 2·25·0.5 / 9.8 ≈ 2.55 s.
(b) R = 25²·sin60° / 9.8 = 625·0.8660 / 9.8 ≈ 55.3 m.
(c) H = 25²·sin²30° / (2·9.8) = 625·0.25 / 19.6 ≈ 7.97 m.

Connecting the Concepts

All the topics discussed are inter‑related:

• Instantaneous velocity and acceleration provide the fundamental definitions from which the equations of motion arise.
• Relative velocity lets us shift between frames—a crucial step when analyzing projectile motion from a moving platform (e.g., a ship or a moving vehicle).
• The graphical treatment of motion offers visual intuition that underpins the algebraic formulas used for free fall and projectile problems.
• Free‑fall motion is a special case of projectile motion with zero launch angle and initial vertical velocity.
• Projectile motion combines uniform horizontal motion (constant velocity) with uniformly accelerated vertical motion (free fall), illustrating the superposition principle.

Mastery of these ideas enables the solution of a wide range of real‑world problems, from sports analytics to engineering design.