AP Physics 1 Unit 1 Notes: Two-Dimensional Kinematics (Algebra-Based)

Reference Frames and Relative Motion

What a reference frame is (and why you can’t avoid choosing one)

A reference frame is the viewpoint you use to measure position, time, and motion. In practice, it’s a coordinate system (like an $x$-axis and $y$-axis) plus an implicit “observer” who considers themselves at rest. Motion is not something an object “has” by itself; motion is a relationship between an object and a chosen frame.

This matters because the same event can produce different measured velocities and positions depending on the observer. If you’re standing on the sidewalk, you measure a passing bus as moving fast. If you’re sitting inside the bus, you measure the passenger next to you as (approximately) not moving at all. Both descriptions can be correct—because they are made in different frames.

In AP Physics 1, you typically treat everyday frames (ground, a moving cart, a boat) as inertial (non-accelerating) unless stated otherwise. In an inertial frame, Newton’s laws and the constant-acceleration kinematics you already know behave “normally.” If the frame itself accelerates, motion descriptions become more complicated (and AP Physics 1 generally avoids heavy use of non-inertial frames in early kinematics).

Relative position and relative velocity (the core idea)

When you switch frames, positions and velocities change by a simple, consistent rule: they add like vectors.

If object $A$ is moving relative to object $B$, and $B$ is moving relative to $C$, then the velocity of $A$ relative to $C$ is the vector sum

$\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$

The notation $A/C$ reads “A relative to C.” This equation is powerful because it turns many “moving observer” problems into organized bookkeeping.

A similar relationship applies to displacement over the same time interval:

$\Delta \vec{r}_{A/C} = \Delta \vec{r}_{A/B} + \Delta \vec{r}_{B/C}$

In AP Physics 1, the most common use is velocity addition. The key is that these are vector equations, not just magnitudes. Direction matters.

How to use relative motion in problem solving

1. Choose a frame (often the ground) and define positive directions.
2. Translate words into relative-velocity statements using clear subscripts.
3. Add vectors using components (usually easiest), especially if directions aren’t collinear.
4. Interpret the result: the final velocity must make physical sense (direction, size, units).

A common misconception is to treat relative velocity as a special “formula to memorize.” It’s really just a consequence of consistency: all observers must agree on how positions relate, even if they measure different positions.

Relative motion in one dimension (warm-up)

When everything is along a line, the vector addition reduces to signed numbers. For example, if a person walks forward at $+1.5\ \text{m/s}$ relative to a train, and the train moves at $+20\ \text{m/s}$ relative to the ground, then the person’s speed relative to ground is

$v_{P/G} = v_{P/T} + v_{T/G}$

The sign convention (what counts as positive) is what keeps the equation honest.

Relative motion in two dimensions (where vectors become essential)

Many real situations involve perpendicular directions: a boat aims across a river while the current carries it downstream, or an airplane points one way while wind pushes another.

In 2D, you almost always want to use components. If

$\vec{v} = \langle v_x, v_y \rangle$

then the relative-velocity equation becomes two separate component equations:

$v_{A/C,x} = v_{A/B,x} + v_{B/C,x}$

$v_{A/C,y} = v_{A/B,y} + v_{B/C,y}$

This is the biggest “unlock” in relative motion: once you break vectors into components, the problem becomes two simpler problems—one in $x$ and one in $y$.

Worked example 1: Boat crossing a river (classic relative velocity)

Situation: A river flows east at $2.0\ \text{m/s}$ relative to the ground. A boat can move at $3.0\ \text{m/s}$ relative to the water, and the pilot points the boat due north (straight across).

Goal: Find the boat’s velocity relative to the ground (magnitude and direction).

Step 1: Write the relative-velocity relationship
Let $B$ be boat, $W$ water, $G$ ground.

$\vec{v}_{B/G} = \vec{v}_{B/W} + \vec{v}_{W/G}$

Step 2: Assign components
Take $+x$ east, $+y$ north.

• Boat relative to water: $\vec{v}_{B/W} = \langle 0, 3.0 \rangle\ \text{m/s}$
• Water relative to ground: $\vec{v}_{W/G} = \langle 2.0, 0 \rangle\ \text{m/s}$

So

$\vec{v}_{B/G} = \langle 2.0, 3.0 \rangle\ \text{m/s}$

Step 3: Magnitude and direction

$v_{B/G} = \sqrt{(2.0)^2 + (3.0)^2} = \sqrt{13}\ \text{m/s} \approx 3.6\ \text{m/s}$

Direction measured east of north:

$\theta = \tan^{-1}\left(\frac{2.0}{3.0}\right) \approx 34^\circ$

So the boat moves at about $3.6\ \text{m/s}$, drifting about $34^\circ$ east of north.

What commonly goes wrong: Students often add magnitudes $3.0 + 2.0 = 5.0$, which ignores perpendicular directions. In 2D, magnitudes only add directly when vectors are collinear.

Worked example 2: “Walking on a moving walkway” (signs and frames)

Situation: A moving walkway carries you west at $1.2\ \text{m/s}$ relative to the ground. You walk east at $1.8\ \text{m/s}$ relative to the walkway.

Take east as positive. Then

$v_{W/G} = -1.2\ \text{m/s}$

$v_{P/W} = +1.8\ \text{m/s}$

Relative velocity:

$v_{P/G} = v_{P/W} + v_{W/G} = 1.8 + (-1.2) = 0.6\ \text{m/s}$

You still move east relative to the ground, but slowly. The main skill here is consistent signs and clear “who relative to whom” subscripts.

Exam Focus

• Typical question patterns:
  • “A boat/plane moves at speed $v$ relative to water/air while current/wind moves at speed $u$ relative to ground; find the velocity relative to ground.”
  • “An observer on a moving cart measures a ball’s velocity; convert to the ground frame (or vice versa).”
  • “Find the required heading so that the resultant velocity points straight across (solve for a component to cancel another).”
• Common mistakes:
  • Adding speeds as scalars when the directions are not the same; fix this by adding components.
  • Mixing up subscripts (e.g., using $\vec{v}_{G/W}$ when you meant $\vec{v}_{W/G}$). Remember reversing the order reverses direction.
  • Forgetting that “relative to” means you may need a vector subtraction; rearrange the relative-velocity equation carefully.

Vectors and Motion in Two Dimensions

What vectors are (and why kinematics becomes cleaner with them)

A vector is a quantity with both magnitude and direction (examples: displacement, velocity, acceleration, force). A scalar has only magnitude (examples: time, mass, temperature, speed).

Two-dimensional motion is hard to describe using only magnitudes because direction changes matter. Vectors let you represent direction in a way that can be calculated, not just described qualitatively.

In kinematics, the central idea is this: if acceleration is constant (or known), you can predict how velocity and position change in time. In 2D, you do the same thing, but separately in each perpendicular direction.

Vector representations you’ll see

You may see vectors written in several equivalent forms:

Two more useful relationships connect component form to magnitude and direction (angle $\theta$ measured from +$x$):

$A = \sqrt{A_x^2 + A_y^2}$

$A_x = A\cos\theta$

$A_y = A\sin\theta$

Here, $A$ could be the magnitude of displacement, velocity, or acceleration, and $A_x$, $A_y$ are its components.

Why components matter: Once you have $x$ and $y$ components, you can apply 1D kinematics independently in each direction. This is the foundation of projectile motion and many AP problems.

Adding and subtracting vectors (the operational skill)

If

$\vec{A} = \langle A_x, A_y \rangle$

$\vec{B} = \langle B_x, B_y \rangle$

then

$\vec{A} + \vec{B} = \langle A_x + B_x, A_y + B_y \rangle$

and

$\vec{A} - \vec{B} = \langle A_x - B_x, A_y - B_y \rangle$

This is why component form is so convenient: vector arithmetic becomes ordinary algebra.

A common mistake is to subtract magnitudes when asked for a vector difference. Always subtract component-by-component, then find magnitude if needed.

Kinematics in 2D: the “separate axes” principle

For many AP Physics 1 situations (especially projectile motion near Earth), acceleration is constant and points straight down:

$\vec{a} = \langle 0, -g \rangle$

with $g$ about $9.8\ \text{m/s}^2$ near Earth’s surface.

That means:

• Horizontally, $a_x = 0$, so horizontal velocity is constant.
• Vertically, $a_y = -g$, so vertical motion is standard constant-acceleration motion.

You apply the constant-acceleration kinematic equations in each direction separately. For a time interval $t$:

$v_x = v_{0x} + a_x t$

$x = x_0 + v_{0x} t + \frac{1}{2}a_x t^2$

$v_y = v_{0y} + a_y t$

$y = y_0 + v_{0y} t + \frac{1}{2}a_y t^2$

Important conceptual point: The motions in $x$ and $y$ are linked by sharing the same time $t$, but otherwise they evolve independently under their respective accelerations.

Projectile motion (a special case of 2D constant-acceleration motion)

A projectile is an object that moves under the influence of gravity alone (neglecting air resistance). In AP Physics 1 problems, that usually means:

$a_x = 0$

$a_y = -g$

The big idea is that gravity affects only the vertical component of motion; it does not “wear down” horizontal velocity (again, only if air resistance is neglected).

To start a projectile problem, you typically:

1. Choose axes (often $+x$ horizontal, $+y$ upward).
2. Break the initial velocity into components.
3. Use vertical motion to determine time or vertical speed.
4. Use that same time in horizontal motion to get range or horizontal displacement.

Worked example 1: Launch at an angle from level ground (range via time)

Situation: A ball is launched from ground level with speed $v_0 = 20\ \text{m/s}$ at angle $\theta = 30^\circ$ above horizontal. Neglect air resistance. Find the time of flight and horizontal range.

Step 1: Decompose the initial velocity

$v_{0x} = v_0\cos\theta = 20\cos 30^\circ \approx 17.3\ \text{m/s}$

$v_{0y} = v_0\sin\theta = 20\sin 30^\circ = 10\ \text{m/s}$

Step 2: Use vertical motion to find total time
Because it lands at the same height, you can use symmetry: time up equals time down. The time to reach the top occurs when $v_y = 0$:

$0 = v_{0y} - g t_{up}$

$t_{up} = \frac{v_{0y}}{g} = \frac{10}{9.8} \approx 1.02\ \text{s}$

Total flight time:

$t = 2t_{up} \approx 2.04\ \text{s}$

Step 3: Use horizontal motion for range
With $a_x = 0$, horizontal displacement is

$\Delta x = v_{0x} t \approx 17.3 \times 2.04 \approx 35.3\ \text{m}$

What commonly goes wrong:

• Plugging $v_0$ into horizontal equations instead of $v_{0x}$.
• Using $g = 9.8$ but forgetting the sign; with $+y$ up, $a_y = -9.8\ \text{m/s}^2$.
• Assuming “range formula” without understanding; AP often designs problems where memorized shortcuts fail (different launch/landing heights).

Worked example 2: Launch from a height (time comes from vertical displacement)

Situation: A ball rolls off a table of height $1.2\ \text{m}$ with horizontal speed $2.5\ \text{m/s}$. Find how far from the table it lands.

Step 1: Identify initial components
The initial vertical velocity is zero because it rolls off horizontally:

$v_{0x} = 2.5\ \text{m/s}$

$v_{0y} = 0$

Also, $a_x = 0$ and $a_y = -g$.

Step 2: Solve for time using vertical motion
Take the table top as $y_0 = 0$ and the ground as $y = -1.2\ \text{m}$. Then

$y = y_0 + v_{0y} t + \frac{1}{2}a_y t^2$

$-1.2 = 0 + 0 + \frac{1}{2}(-9.8)t^2$

$t^2 = \frac{2(1.2)}{9.8} \approx 0.245$

$t \approx 0.495\ \text{s}$

Step 3: Use time in horizontal motion

$\Delta x = v_{0x} t \approx 2.5 \times 0.495 \approx 1.24\ \text{m}$

What commonly goes wrong: Students try to use a “horizontal range” equation from level-ground launches. When heights differ, you must use vertical displacement to find time.

Velocity and acceleration vectors: what they mean physically

In 2D motion, it’s helpful to remember what the vectors “point along”:

• The velocity vector points in the direction of motion at that instant (tangent to the path).
• The acceleration vector points in the direction the velocity is changing.

For a projectile (no air resistance), acceleration always points downward, even at the top of the trajectory. A classic misconception is that acceleration becomes zero at the peak because vertical velocity is zero. The velocity can be zero in one direction while acceleration is still nonzero.

You can also connect components back to the overall speed:

$v = \sqrt{v_x^2 + v_y^2}$

This is often used when AP asks for “speed at a point” after finding components.

Interpreting motion graphs in 2D contexts

AP Physics 1 often uses graphs to test whether you understand components and slopes/areas.

• On a $v_x$ vs. $t$ graph for projectile motion, you expect a horizontal line (constant $v_x$).
• On a $v_y$ vs. $t$ graph, you expect a straight line with slope $-g$.
• On a $y$ vs. $t$ graph, you expect a concave-down parabola.

Even if a problem is “2D,” many graph questions isolate one component at a time.

Putting it together: a structured method for any 2D kinematics problem

When you feel lost, return to a repeatable approach:

1. Sketch the motion and choose axes.
2. List knowns in $x$ and in $y$ separately (including signs).
3. Write equations for $x(t)$ and $y(t)$ (or velocity forms) using the correct accelerations.
4. Use time as the link between directions.
5. Check reasonableness: units, signs, limiting cases.

A helpful memory aid is: “Same time, separate equations.” The time interval $t$ is shared; everything else can be treated independently per axis.

Worked example 3: Find speed and direction of velocity at a time

Situation: A projectile is launched with $v_0 = 15\ \text{m/s}$ at $40^\circ$ above horizontal. Find the velocity (components, speed, and direction) at $t = 1.0\ \text{s}$.

Step 1: Initial components

$v_{0x} = 15\cos 40^\circ \approx 11.5\ \text{m/s}$

$v_{0y} = 15\sin 40^\circ \approx 9.64\ \text{m/s}$

Step 2: Update components using acceleration
Horizontal:

$v_x = v_{0x} + a_x t = 11.5 + 0 = 11.5\ \text{m/s}$

Vertical:

$v_y = v_{0y} + a_y t = 9.64 - 9.8(1.0) \approx -0.16\ \text{m/s}$

Step 3: Speed

$v = \sqrt{(11.5)^2 + (-0.16)^2} \approx 11.5\ \text{m/s}$

Step 4: Direction

$\phi = \tan^{-1}\left(\frac{v_y}{v_x}\right) \approx \tan^{-1}\left(\frac{-0.16}{11.5}\right) \approx -0.8^\circ$

So at $t = 1.0\ \text{s}$, it’s moving almost horizontally, slightly downward.

What commonly goes wrong:

• Using speed instead of velocity components in the kinematic update.
• Assuming the projectile must still be moving upward at $t = 1.0\ \text{s}$ without checking the numbers.

Exam Focus

• Typical question patterns:
  • “A projectile is launched with speed and angle; find time of flight, maximum height, range, or velocity at a given time.”
  • “An object is launched horizontally from a height; find landing distance or impact speed.”
  • “Given a vector (or a diagram), find components, then use kinematics separately in $x$ and $y$; recombine to get magnitude/direction.”
• Common mistakes:
  • Treating $v_x$ as changing due to gravity (it doesn’t if air resistance is neglected); keep $a_x = 0$.
  • Setting acceleration to zero at the top of a trajectory; only $v_y$ is zero there, not $a_y$.
  • Mixing up angles: using $\sin$ and $\cos$ incorrectly for components. A reliable check is: at small angles, $v_{0x}$ should be close to $v_0$ and $v_{0y}$ should be smaller.