Kinematics Formulas to Know for AP Physics 1

What You Need to Know

Kinematics is the math of motion (position, velocity, acceleration) without worrying about forces. In AP Physics 1, most kinematics questions boil down to: choose a coordinate system, translate words/graphs into variables, then use the right constant-acceleration or 2D component equations.

Core quantities (with meanings)

• Position $x$ (1D) or $\vec r$ (2D): where you are.
• Displacement $\Delta x = x - x_0$: change in position.
• Velocity $v$ or $\vec v$: how fast and which direction your position changes.
• Acceleration $a$ or $\vec a$: how fast and which direction your velocity changes.

The big idea you must internalize

If acceleration is constant, the “Big 4” kinematics equations apply. If acceleration is not constant, you rely on graphs, areas, and definitions (still algebra-based).

Constant acceleration is the default AP Physics 1 kinematics model

• “Free fall” near Earth: $a_y = -g$ where $g \approx 9.8\ \mathrm{m/s^2}$ (often rounded to $10\ \mathrm{m/s^2}$), direction downward.
• Projectile motion (no air resistance):
  • $a_x = 0$
  • $a_y = -g$
  • Motion in $x$ and $y$ are linked only by **time** $t$.

Step-by-Step Breakdown

Use this every time you see a kinematics problem.

1. Choose a coordinate system and write sign conventions

   • Pick +x or +y directions intentionally (often up is +y).
   • Write what is positive/negative before plugging numbers.

2. List knowns/unknowns (by component if 2D)
   Typical variables:

   • $x_0,\ x,\ \Delta x$
   • $v_0,\ v$
   • $a$
   • $t$

3. Decide: constant acceleration or not?

   • If constant acceleration is stated or implied (free fall, constant braking, constant thrust): use the Big 4.
   • If you’re given a graph $v(t)$ or $a(t)$ and it’s piecewise: use slope/area relationships.

4. If motion is 2D, split into components

   • $v_{0x} = v_0\cos\theta$
   • $v_{0y} = v_0\sin\theta$
   • Solve $x$-direction and $y$-direction separately, then recombine if needed.

5. Pick the equation that avoids your “extra” unknown
   Example decision point:

   • Need $v$ but not $t$? Use $v^2 = v_0^2 + 2a\Delta x$.
   • Need $\Delta x$ and have $t$? Use $\Delta x = v_0 t + \tfrac{1}{2}at^2$.

6. Check units + reasonableness

   • Times should be positive.
   • Speeds shouldn’t come out negative (negative belongs to velocity components).
   • In free fall, if up is positive, then $a_y$ should be negative.

Micro-worked example (method in action)

A ball is thrown upward with $v_{0y} = +12\ \mathrm{m/s}$ from $y_0 = 0$. Find time to peak.
1) Choose up as +y, so $a_y = -9.8\ \mathrm{m/s^2}$.
2) At peak, $v_y = 0$.
3) Use $v_y = v_{0y} + a_y t$:

$0 = 12 + (-9.8)t \Rightarrow t \approx 1.22\ \mathrm{s}$

Key Formulas, Rules & Facts

Definitions (always true)

Constant-acceleration (1D) “Big 4”

Assume $a$ is constant.

Free-fall shortcuts (vertical motion)

Take up as +y (common choice):

• $a_y = -g$
• $v_y = v_{0y} - gt$
• $\Delta y = v_{0y} t - \tfrac{1}{2}gt^2$
• $v_y^2 = v_{0y}^2 - 2g\Delta y$

Facts you use all the time:

• At the top of a vertical toss: $v_y = 0$ (but $a_y$ is still $-g$).
• If it returns to the same height (no air resistance): time up = time down, and $|v_{\text{landing}}| = |v_{\text{launch}}|$ (direction flips).

Projectile motion (2D, no air resistance)

Split into independent components.

Vector magnitude relationships:

• Speed: $v = \sqrt{v_x^2 + v_y^2}$
• Launch components: $v_{0x} = v_0\cos\theta$, $v_{0y} = v_0\sin\theta$

Special results (only when launch and landing heights match):

• Time of flight: $T = \dfrac{2v_0\sin\theta}{g}$
• Range: $R = \dfrac{v_0^2\sin(2\theta)}{g}$
• Max height above launch: $H = \dfrac{v_0^2\sin^2\theta}{2g}$

If launch and landing heights are different, do not use the “same-height” shortcut formulas. Use component kinematics with $y(t)$ instead.

Motion graphs (the algebra-based superpower)

For straight-line segments on graphs:

• Average velocity on a time interval: $v_{\text{avg}} = \dfrac{\Delta x}{\Delta t}$
• For a linear $v(t)$ segment: area is triangle/trapezoid geometry.

Relative motion (often a quick plug-in)

• Relative velocity in 1D: $v_{A/B} = v_A - v_B$
• Relative position: $x_{A/B} = x_A - x_B$

Uniform circular motion (kinematics of turning)

AP Physics 1 often treats this as kinematics + dynamics later. Formulas you should know:

Examples & Applications

Example 1: 1D constant acceleration (braking)

A car at $v_0 = 20\ \mathrm{m/s}$ brakes with constant $a = -4\ \mathrm{m/s^2}$. How far to stop?

• Stopping means $v = 0$.
• Use the no-time equation: $v^2 = v_0^2 + 2a\Delta x$.

$0 = (20)^2 + 2(-4)\Delta x \Rightarrow \Delta x = 50\ \mathrm{m}$

Key insight: when time isn’t asked for (and not given), jump to $v^2 = v_0^2 + 2a\Delta x$.

Example 2: Vertical toss (max height)

Throw upward with $v_{0y} = +14\ \mathrm{m/s}$. Find max height above launch.

• At top: $v_y = 0$.
• Use: $v_y^2 = v_{0y}^2 + 2a_y\Delta y$ with $a_y = -g$.

$0 = (14)^2 + 2(-9.8)\Delta y \Rightarrow \Delta y = \dfrac{196}{19.6} = 10\ \mathrm{m}$

Key insight: max height problems are almost always “set $v=0$ at the top.”

Example 3: Projectile launched from a height (no same-height shortcuts)

A ball is launched from $y_0 = 5\ \mathrm{m}$ with $v_0 = 10\ \mathrm{m/s}$ at $\theta = 30^\circ$. Find time to hit the ground.

• Compute components: $v_{0y} = 10\sin 30^\circ = 5\ \mathrm{m/s}$.
• Vertical equation with ground at $y = 0$:

$0 = 5 + (5)t - \tfrac{1}{2}(9.8)t^2$

Solve quadratic; you’ll get two solutions, take the positive landing time (the other is unphysical).

Key insight: different start/end heights forces you into $y(t)$ and a quadratic.

Example 4: Read a $v$–$t$ graph via area

If a particle has:

• $v = 2\ \mathrm{m/s}$ from $t=0$ to $t=3\ \mathrm{s}$,
• then increases linearly to $v = 6\ \mathrm{m/s}$ by $t=5\ \mathrm{s}$,
  then displacement is area under $v(t)$:
• Rectangle: $\Delta x_1 = (2)(3) = 6\ \mathrm{m}$
• Trapezoid from 3 to 5 s: average velocity $= \tfrac{2+6}{2} = 4\ \mathrm{m/s}$, so $\Delta x_2 = (4)(2) = 8\ \mathrm{m}$
• Total: $\Delta x = 14\ \mathrm{m}$

Key insight: for linear segments, treat areas as rectangles/triangles/trapezoids.

Common Mistakes & Traps

1. Mixing up distance and displacement

   • Wrong: using total path length in $v_{\text{avg}} = \Delta x/\Delta t$.
   • Why wrong: $\Delta x$ is straight-line change in position (with sign).
   • Fix: compute displacement from initial and final positions.

2. Forgetting velocity and acceleration are vectors (sign matters)

   • Wrong: plugging $a = 9.8$ when you chose up as positive.
   • Why wrong: in that coordinate, $a_y = -9.8\ \mathrm{m/s^2}$.
   • Fix: write your sign convention first; attach signs to every component.

3. Assuming $v = 0$ means $a = 0$

   • Wrong: saying acceleration is zero at the top of a toss.
   • Why wrong: gravity is still acting, so $a_y = -g$ at all times (ignoring air).
   • Fix: only set $v_y = 0$ at the peak, not acceleration.

4. Using same-height projectile formulas when heights differ

   • Wrong: using $T = \dfrac{2v_0\sin\theta}{g}$ when launched from a cliff.
   • Why wrong: that formula assumes landing height equals launch height.
   • Fix: use $y = y_0 + v_{0y}t - \tfrac{1}{2}gt^2$ with the actual final $y$.

5. Component confusion in projectile motion

   • Wrong: letting $v_x$ change because $v$ changes.
   • Why wrong: with no air resistance, $a_x = 0$ so $v_x$ stays constant.
   • Fix: treat $x$ and $y$ separately; only $v_y$ changes with gravity.

6. Picking the wrong time (multiple solutions)

   • Wrong: taking the negative root from a quadratic in $t$.
   • Why wrong: negative time usually corresponds to an earlier crossing before your chosen $t=0$.
   • Fix: interpret roots physically; choose the time consistent with the scenario.

7. Sign errors in $v^2 = v_0^2 + 2a\Delta x$

   • Wrong: plugging $\Delta x$ as a positive magnitude automatically.
   • Why wrong: $\Delta x$ can be negative depending on your axis direction.
   • Fix: compute $\Delta x = x - x_0$ with sign.

8. Misreading graph area vs slope

   • Wrong: treating area under $v(t)$ as acceleration.
   • Why wrong: slope gives acceleration; area gives displacement.
   • Fix: memorize: “slope tells the derivative; area tells the accumulation.”

Memory Aids & Quick Tricks

Quick Review Checklist

• You can write the definitions: $v_{\text{avg}} = \Delta x/\Delta t$ and $a_{\text{avg}} = \Delta v/\Delta t$.
• You know the Big 4 constant-acceleration equations and when each is best.
• You always choose an axis and keep signs consistent (especially for free fall).
• For vertical motion, you automatically set $a_y = -g$ (if up is positive).
• For projectiles: $a_x=0$ so $v_x$ is constant; $y(t)$ controls flight time.
• You can extract motion from graphs: slope vs area rules.
• You avoid same-height shortcuts unless the start and end heights truly match.
• You sanity-check units and physical direction (negative velocity is fine; negative speed isn’t).

You’ve got this—most kinematics points come from clean setup, consistent signs, and the right equation choice.