A systematic review of kinematic equations — from constant velocity through free-fall projectile motion.
Chapter 01
Speed & Velocity
Speed is a scalar — it tells you how fast. Velocity is a vector — it tells you how fast and in which direction. This distinction becomes critical once objects change direction.
Average Speed
s = d / t
Total distance divided by total time. Always positive — no direction involved.
Average Velocity
v̄ = Δx / Δt
Displacement (Δx = xf − xi) divided by elapsed time. Can be negative.
Displacement
Δx = xf − xi
Change in position. A straight-line quantity — not the path length traveled.
Speed vs. Velocity — Key Distinction
If a runner completes a 400 m lap and returns to the start, their displacement = 0, so average velocity = 0 m/s. Their average speed, however, is distance / time and is nonzero. Never substitute one for the other.
Symbol
Quantity
SI Unit
Type
s
Speed
m/s
Scalar
v̄
Average velocity
m/s
Vector
d
Distance
m
Scalar
Δx
Displacement
m
Vector
Δt
Elapsed time
s
Scalar
Chapter 02
Acceleration
Acceleration is the rate of change of velocity. An object accelerates whenever it speeds up, slows down, or changes direction — even at constant speed.
Average Acceleration
ā = Δv / Δt
Change in velocity over elapsed time. Measured in m/s².
Change in Velocity
Δv = vf − vi
Final velocity minus initial velocity. Negative Δv means deceleration in the positive direction.
Gravitational Accel.
g = −9.8 m/s²
Near Earth's surface, downward. Often approximated as −10 m/s² for estimation.
Sign Convention
Define a positive direction first — usually upward or rightward. Then acceleration opposing motion carries a negative sign. A car braking from +20 m/s to rest has Δv = −20 m/s, so acceleration is negative (deceleration).
Worked Example · Acceleration
A car accelerates from rest to 28 m/s in 7 seconds. What is its average acceleration?
01
Identify: vi = 0 m/s, vf = 28 m/s, Δt = 7 s
02
Formula: ā = Δv / Δt = (vf − vi) / Δt
03
Solve: ā = (28 − 0) / 7 = 4 m/s²
Chapter 03
The Big Four
For constant acceleration, four algebraic equations relate displacement, initial/final velocity, acceleration, and time. Together they can solve any 1D kinematics problem if you know three of the five variables.
Equation I — missing Δx
vf = vi + at
Velocity as a function of time and constant acceleration. Use when displacement is not needed.
Equation II — missing vf
Δx = vit + ½at²
Position under constant acceleration. Quadratic in time — may produce two solutions.
Equation III — missing t
vf² = vi² + 2aΔx
The velocity-displacement relation. Time-independent; ideal for projectile peak height.
Equation IV — missing a
Δx = ½(vi + vf)t
Average velocity × time. Only valid under constant acceleration (linear velocity change).
Strategy: Variable Inventory
List the five kinematic variables: Δx, vi, vf, a, t. Mark the three that are known and one unknown. Then pick the equation that contains exactly those four, ignoring the fifth (the missing variable).
Symbol
Quantity
SI Unit
Δx
Displacement
m
vi
Initial velocity
m/s
vf
Final velocity
m/s
a
Constant acceleration
m/s²
t
Time elapsed
s
Chapter 04
Thrown Projectiles
A projectile is any object moving under gravity alone (ignoring air resistance). The key insight: horizontal and vertical motion are completely independent. Treat them as two separate 1D problems that share only the time variable.
Launch (x₀, y₀)← Range (R) →Landing
Horizontal (x-axis)
ax = 0 (no acceleration)
vx = v₀ cos θ (constant)
x = v₀ cos θ · t
Δx = vx · t
Vertical (y-axis)
ay = −g = −9.8 m/s²
vy = v₀ sin θ − gt
y = v₀ sin θ · t − ½gt²
vy² = (v₀ sin θ)² − 2gΔy
Time to Peak
tpeak = viy / g
Set vy = 0. Time at max height is half the total flight time (symmetric launch & landing).
Maximum Height
H = viy² / (2g)
From vy² = viy² − 2gH with vy = 0 at peak.
Total Range
R = vx · ttotal
Horizontal velocity × total flight time. For flat ground: ttotal = 2 · tpeak.
Launch Decomposition
v0x = v₀ cos θ v0y = v₀ sin θ
Decompose initial speed v₀ at angle θ above horizontal into components.
At the Peak — vy = 0
At maximum height, vertical velocity is zero for an instant. Horizontal velocity never changes. So total speed at the peak equals vx = v₀ cos θ, not zero.
Worked Example · Projectile
A ball is kicked at 20 m/s at 30° above horizontal from ground level. Find: (a) max height, (b) total range.
01
Decompose: v0x = 20 cos 30° ≈ 17.3 m/s | v0y = 20 sin 30° = 10 m/s