Kinematics: Motion, Vectors, and Equations for Physics

Kinematics

Study of motion without considering the causes (forces).

Position

Location of an object in space relative to a reference coordinate system (e.g., 1‑D line, 2‑D Cartesian).

Displacement (Δx)

Net change in position: x_f − x_0. Direction matters.

Distance

Total path length traveled; always nonnegative and path‑dependent.

Speed

Rate of change of position with respect to time; scalar (no direction).

Velocity

Rate of change of position with respect to time; vector (includes direction).

Acceleration

Rate of change of velocity with respect to time; vector.

Reference frame / coordinate system

The chosen spatial reference (e.g., number line, Cartesian axes) that defines directions and origins for vectors.

Time in kinematics

Independent parameter needed to define motion; treated separately from spatial coordinates here.

Scalar

Quantity with magnitude only (e.g., distance, speed).

Vector

Quantity with magnitude and direction, represented by components along coordinate axes.

Velocity vs. Speed

Velocity includes sign/direction; speed is the magnitude of velocity.

Distance vs. Displacement

Distance sums all movement; displacement is the straight‑line change from start to finish and can be zero even if distance > 0.

When a and v differ in direction

Turning on a curve or braking while moving forward—acceleration vector not aligned with velocity.

Reporting position in daily life

We implicitly choose references (addresses, GPS lat/long, campus map coordinates).

Average velocity (v_avg)

v_avg = (x_f − x_0)/(t_f − t_0); sign shows direction in chosen axis.

Instantaneous speed

The instantaneous rate of change of position with time (magnitude only).

Average acceleration (a_avg)

a_avg = (v_f − v_0)/(t_f − t_0).

Slope links (P→V)

Slope of position-time graph equals velocity.

Slope links (V→A)

Slope of velocity-time graph equals acceleration.

Trinary operations (P, V, A)

Use slope between P/V/A plots to move 'down' (P→V, V→A) or integrate conceptually to move 'up'. Work piecewise when behavior changes.

Cyclist: displacement

3E, 5W, 1E, 9W ⇒ Δx = −10 km (10 km West).

Cyclist: total distance

3 + 5 + 1 + 9 = 18 km.

Example: average velocity sign

From x0=8 to xf=4 in 2 s ⇒ v_avg = (4−8)/2 = −2 units/s (negative direction).

Acceleration need not change speed

A curved path can have acceleration from changing direction even at constant speed.

Reasonable assumptions

If the reference frame isn't stated, pick one and state it.

Constant‑a kinematics: v

v_f = v_0 + a t (for constant acceleration).

Constant‑a kinematics: x

x_f = x_0 + v_0 t + (1/2) a t^2.

Constant‑a kinematics: v^2

v_f^2 = v_0^2 + 2 a (x_f − x_0).

Constant‑a kinematics: average v

For constant a: v_avg = (v_0 + v_f)/2.

Condition for use

These equations require constant (uniform) acceleration.

Zero acceleration case

Position vs. time is linear; velocity is constant; acceleration is zero.

Meeting time (example)

Two objects' position plots cross at t_c = 7 in a sample case.

Meeting time (another example)

From simple V→P plotting, a crossover occurs at t_c = 3.5 in a sample case.

Initial position to meet later

Choose x0 so positions are equal at a target time; e.g., x0 = −5 (sample).

Initial positions to meet later

Another sample: x0 = (5, −10) for two objects to meet at t=10.

0→60 mph in 8.5 s: a

a ≈ 3.14 m/s^2 (convert mph→m/s, then Δv/Δt).

0→60 mph: distance

x − x0 ≈ 113 m ≈ 126 yd (with constant a).

'Texting at 150 km/h' takeaway

Braking with earlier a won't stop in 50 yd; impact ≈ 84.8 mph in that scenario.

Gravity in 1‑D (vertical motion)

Use same constant‑a equations with a = −g (take g ≈ 9.80 m/s^2 downward).

Parabolic position‑time

Under constant acceleration, position vs. time is quadratic (parabola).

Free‑fall from rest

x − x0 = (1/2) g t^2 downward; time controls distance quadratically.

Toss up then down

Velocity changes sign at the peak (v=0), position is max there; motion is symmetric (ignoring air).

Lamborghini GT3 Evo 2: a

Accelerates to 300 km/h in 7.5 s ⇒ a ≈ 11.11 m/s^2 (~1.13 g).

Lamborghini: braking

Stops in 3.2 s from 300 km/h ⇒ a ≈ −26 m/s^2 (~−2.66 g).

Lamborghini: total distance

About 868 m across accelerate-cruise-brake sequence.

Ball toss: v0=30 m/s @ 1.5 m

Peak at ~47.5 m, time to peak ~3 s.

Drop: hits in 11 s

Height ≈ 593 m (ignoring air resistance).

Vectors: definition

Array of components (e.g., ⟨x,y⟩) representing magnitude and direction in a coordinate system.

Vector magnitude |v|

|v| = sqrt(v_x^2 + v_y^2).

Unit/normalized vector

û = v / |v|; has length 1 and gives direction.

Direction by angle θ

Angle measured from +x axis; components v_x = |v| cosθ, v_y = |v| sinθ.

Vector addition (analytic)

Add component‑wise: ⟨a_x + b_x, a_y + b_y⟩.

Vector addition (graphical)

Head‑to‑tail construction; resultant is from start to final head.

Example E=⟨13,9⟩ magnitude

|E| ≈ 16; unit ≈ ⟨0.82, 0.57⟩; angle ≈ 35°.

Convert mag‑angle → components

Given |v|,θ: v_x = |v|cosθ, v_y = |v|sinθ.

Vector operations: recap

Add/subtract vectors by components or head‑to‑tail; scalars scale magnitude (and flip direction if negative).

Magnitude from components

|v| = sqrt(v_x^2 + v_y^2).

Unit/normalized direction

û = v/|v| gives direction with length 1.

Angle representation

Describe direction by angle from +x; convert with cos/sin.

Projectile: level‑to‑level setup

Split initial velocity into v0x, v0y; solve horizontal and vertical 1‑D motions separately (a_x=0, a_y=−g).

Baseball HR scenario

Exit 56 m/s at 20°, 1 m high; 3 m fence at 150 m → clears fence (Yes).

Vertical launch v0=29 m/s @ 3 m

Find peak height/time and total time by 1‑D constant‑a with a=−g.

Incline roll problem

Decompose g along slope: a = g sin(θ) downhill (no friction assumed) to find time and speed.

Chase & stop time thought

Compare car's downhill time to runner speed (5 m/s) to estimate if intervention is possible.