Kinematics
1. Position, Velocity, and Acceleration
Kinematics is the description of motion without asking what causes it. Before you can answer "why does it move that way," you have to be able to answer "how is it moving, exactly" — and that means being precise about three quantities: where something is, how fast it's changing position, and how fast that rate of change is itself changing.
The setup
xposition (m) — where the object is along a chosen axis, relative to a chosen origin.vvelocity (m/s) — the rate of change of position. A vector: direction matters.aacceleration (m/s²) — the rate of change of velocity. Also a vector.ttime (s) — measured from some reference point, usuallyt = 0at the start of the motion you're analyzing.
The most important habit to build here: velocity tells you nothing about speeding up or slowing down, and acceleration tells you nothing about which way the object is currently moving. They are independent pieces of information. An object can have negative velocity and positive acceleration at the same instant (slowing down while moving in the negative direction) — that's not a contradiction, it's just two separate facts.
From position to velocity to acceleration
Velocity is the slope of the position-vs-time graph. Acceleration is the slope of the velocity-vs-time graph. This is the calculus underneath kinematics even before you formally take a derivative:
a = Δv / Δt (average acceleration)
When Δt shrinks toward zero, these become instantaneous velocity and instantaneous acceleration — the actual slope of the curve at one specific point in time, not just the average over an interval.
2. The Kinematic Equations
These four equations only apply when acceleration is constant. That restriction is the single most-overlooked detail in this unit — if acceleration is changing, none of these are valid, and you need calculus (or a different unit's tools) instead.
Notice each equation is missing a different variable. That's the actual skill in this section: not memorizing four formulas, but reading a problem, listing what you know and what you want, and picking the equation that doesn't require a variable you don't have.
A car starts at rest and accelerates at 3.0 m/s² for 6.0 seconds. How far does it travel?
- List knowns: v₀ = 0 m/s, a = 3.0 m/s², t = 6.0 s. Unknown: Δx.
- No final velocity given and none asked for — so use the equation that doesn't involve v: x = x₀ + v₀t + ½at².
- Substitute: x = 0 + (0)(6.0) + ½(3.0)(6.0)².
- Compute: ½(3.0)(36) = 54.
- The car travels 54 meters.
"If acceleration is zero, the object isn't moving." False — zero acceleration means constant velocity, which could be a large constant speed in a straight line. Acceleration measures change, not motion itself.
3. Free Fall
Free fall is just the kinematic equations with a fixed value for acceleration: gravity near Earth's surface, g = 9.8 m/s², always directed downward, regardless of whether the object is moving up, down, or momentarily at rest at the top of its path.
"At the top of its path, a thrown ball has zero acceleration." False — velocity is zero there, not acceleration. Gravity never switches off; the ball's acceleration is −9.8 m/s² the entire time it's in the air, including at the peak.
A ball is thrown straight up at 14.7 m/s. How long until it returns to its starting height?
- Take up as positive. v₀ = 14.7 m/s, a = −9.8 m/s². It returns to the same height, so Δx = 0.
- Use x = x₀ + v₀t + ½at² with Δx = 0: 0 = v₀t + ½at².
- Factor out t: 0 = t(v₀ + ½at). One solution is t = 0 (the start); the other is t = −2v₀/a.
- t = −2(14.7)/(−9.8) = 3.0 seconds.
4. Projectile Motion
Projectile motion is two completely independent 1-D kinematics problems happening at the same time: horizontal motion with zero acceleration, and vertical motion with constant acceleration g. They don't affect each other. This independence is the entire concept — once you accept it, projectile motion is just two copies of what you already did in sections 1–3.
The setup
vₓhorizontal velocity — constant throughout the flight (no horizontal acceleration, ignoring air resistance).v_yvertical velocity — changes at rate g, exactly like free fall.v₀,θlaunch speed and angle — split into components with vₓ = v₀cos θ and v_y = v₀sin θ.
"A heavier projectile falls faster and lands sooner." False, ignoring air resistance — mass cancels out of every kinematic equation here. A bowling ball and a tennis ball launched identically land at the same time. What actually changes the time of flight is launch angle and speed, never mass.
A ball is launched at 20 m/s at 30° above horizontal from ground level. How far away does it land?
- Components: vₓ = 20cos30° ≈ 17.3 m/s, v_y₀ = 20sin30° = 10 m/s.
- Time of flight (returns to y = 0): using the free-fall result pattern, t = 2v_y₀/g = 2(10)/9.8 ≈ 2.04 s.
- Range: x = vₓ · t = 17.3 × 2.04 ≈ 35.3 meters.
5. Simulator
Predict before you check: pick a launch angle and speed, predict the range, then run it.
6. Practice Problems
1. A car traveling at 20 m/s brakes to a stop over a distance of 50 m. Find its acceleration.
2. A stone is dropped from a 45 m cliff. How long does it take to hit the ground?
3. A projectile is launched horizontally at 15 m/s from a 20 m high platform. How far from the base of the platform does it land?