Unit 1 Kinematics: Motion in Two Dimensions (AP Physics C: Mechanics)

Vectors and Vector Components

Two-dimensional kinematics is really one big idea: motion in the plane can be described by tracking the same physical quantities you already know (position, velocity, acceleration), but now each quantity has both an x-part and a y-part. Vectors are the language that makes that precise.

What a vector is (and why you care)

A vector is a quantity with magnitude and direction. In kinematics, the most important vectors are:

• Displacement \Delta \vec{r} (change in position)
• Velocity \vec{v} (rate of change of position)
• Acceleration \vec{a} (rate of change of velocity)

Scalars (like time t, mass m, speed, distance) only have magnitude.

Why this matters: almost every 2D problem becomes manageable when you replace “direction words” (up-left, down-right, etc.) with components. Once you break a vector into its x and y components, you can apply 1D kinematics separately along each axis and then recombine.

Vector notation and component form

In 2D, you can represent a vector in component form:

\vec{A} = A_x \hat{i} + A_y \hat{j}

• A_x is the x-component (how much of the vector points along the +x direction)
• A_y is the y-component (how much of the vector points along the +y direction)
• \hat{i} and \hat{j} are unit vectors (magnitude 1) pointing along +x and +y

A helpful interpretation: components are like “shadow lengths” on the axes. You can think of A_x and A_y as the signed results of projecting \vec{A} onto the coordinate axes.

Magnitude and direction from components

If you know components, the magnitude is given by the Pythagorean relationship:

|\vec{A}| = \sqrt{A_x^2 + A_y^2}

The direction (angle \theta measured from +x, counterclockwise by convention) is related by:

\tan \theta = \frac{A_y}{A_x}

Be careful: the tangent relation alone can put you in the wrong quadrant. If A_x is negative, the vector points left, and you must adjust the angle accordingly.

Components from magnitude and angle

If a vector has magnitude A and makes angle \theta above the +x axis:

A_x = A \cos \theta

A_y = A \sin \theta

Common sign mistake: students sometimes attach negative signs based on “down” or “left” while also choosing angles that already encode direction. Pick one consistent method: either use a signed angle and standard trig, or use a positive reference angle and assign signs from the quadrant.

Vector addition and subtraction (how motion combines)

Vectors add component-wise:

\vec{C} = \vec{A} + \vec{B}

C_x = A_x + B_x

C_y = A_y + B_y

Subtraction is adding the negative:

\vec{A} - \vec{B} = \vec{A} + (-\vec{B})

This matters in kinematics because displacement adds over multiple legs of motion, and velocities add in relative motion.

Kinematics quantities in vector form

In 2D, the position vector is often written:

\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}

Then velocity and acceleration are time derivatives (AP Physics C is calculus-based, so this viewpoint is fundamental):

\vec{v}(t) = \frac{d\vec{r}}{dt}

\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}

Component-wise, that means:

v_x = \frac{dx}{dt}

v_y = \frac{dy}{dt}

a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}

a_y = \frac{dv_y}{dt} = \frac{d^2y}{dt^2}

A powerful consequence: if acceleration is constant, you can use constant-acceleration kinematics in each direction independently.

Worked example: resolve an initial velocity into components

A ball is launched with speed v_0 = 20\ \text{m/s} at \theta = 30^\circ above horizontal. Find v_{0x} and v_{0y}.

Use the component formulas:

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

v_{0y} = v_0 \sin \theta = 20\sin 30^\circ = 10.0\ \text{m/s}

Interpretation: the launch gives the projectile a steady horizontal “carry” of 17.3\ \text{m/s} (if air resistance is neglected) and an upward initial vertical motion of 10.0\ \text{m/s} that gravity will reduce.

Worked example: add displacement vectors

A hiker walks 300\ \text{m} east, then 400\ \text{m} north. Find the magnitude of the total displacement.

Treat each leg as components:

\Delta x = 300\ \text{m}

\Delta y = 400\ \text{m}

Magnitude:

|\Delta \vec{r}| = \sqrt{300^2 + 400^2} = 500\ \text{m}

A frequent misconception here is to say “distance traveled is 500 m.” That’s incorrect: the distance is 300 + 400 = 700\ \text{m} while the **displacement magnitude** is 500\ \text{m}.

Notation reference (common equivalents)

Exam Focus

• Typical question patterns:
  • “Break this velocity/force/displacement into components and use those to solve for a time, speed, or location.”
  • “Given components, find magnitude and direction (or vice versa).”
  • “Write parametric equations x(t) and y(t) for motion under constant acceleration.”
• Common mistakes:
  • Mixing up sine and cosine because the angle is not measured from the +x axis.
  • Getting the quadrant wrong when computing an angle from \tan \theta = A_y/A_x.
  • Treating vectors like scalars (adding magnitudes instead of adding components).

Projectile Motion

Projectile motion is a special case of 2D motion where the only acceleration is due to gravity (near Earth’s surface) and air resistance is neglected. The key idea is simple but easy to misuse:

Gravity acts vertically, so the horizontal and vertical motions are independent except for sharing the same time variable t.

That independence is why projectile problems often feel like two separate 1D problems stitched together.

The physical model (assumptions you must know)

In AP Physics C projectile motion problems, unless explicitly stated otherwise, you typically assume:

• No air resistance
• Constant gravitational acceleration downward
• Choose +y upward (common convention)

Then the acceleration components are:

a_x = 0

a_y = -g

where g is the magnitude of gravitational acceleration (approximately 9.8\ \text{m/s}^2 near Earth).

Why this matters: if a_x = 0, then v_x is constant. Many errors come from accidentally applying -g in the horizontal direction or letting v_x change without a physical reason.

Building the equations of motion (component-by-component)

Start with an initial position x_0, y_0 and initial velocity components v_{0x}, v_{0y}.

Horizontal motion (constant velocity)

With a_x = 0:

v_x(t) = v_{0x}

x(t) = x_0 + v_{0x} t

Vertical motion (constant acceleration)

With a_y = -g:

v_y(t) = v_{0y} - g t

y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2

These are the backbone equations. Nearly every projectile question is solved by combining them with a condition like “hits the ground” (a particular y value) or “reaches maximum height” (a particular v_y value).

Interpreting the motion (what’s really happening)

It’s useful to narrate the physics:

• The projectile keeps whatever horizontal velocity it starts with because there is no horizontal acceleration.
• Vertically, gravity steadily decreases v_y until it reaches zero at the top of the trajectory, then makes v_y negative on the way down.

A common misconception: “At the top, the velocity is zero.” Not in general. At the top, only the vertical component is zero:

v_y = 0

but v_x is still whatever it was initially.

Key results for level-ground launches (same initial and final height)

If the projectile is launched from height y_0 and lands back at the same height, you can derive convenient expressions.

Time to reach maximum height

At the top, v_y = 0:

0 = v_{0y} - g t_{top}

t_{top} = \frac{v_{0y}}{g}

Maximum height above launch point

Plug t_{top} into the vertical position change:

\Delta y_{max} = \frac{v_{0y}^2}{2g}

This result is especially useful because it does not require you to solve a quadratic.

Total time of flight

For symmetric flight (back to same height):

t_{flight} = \frac{2v_{0y}}{g}

Horizontal range

Range R is the horizontal displacement when it lands:

R = v_{0x} t_{flight}

Using v_{0x} = v_0\cos\theta and v_{0y} = v_0\sin\theta gives:

R = \frac{v_0^2 \sin(2\theta)}{g}

Important: this range formula assumes same launch and landing height and no air resistance.

Projectiles launched from a height (most realistic problems)

If the projectile lands at a different height than where it was launched, the motion is no longer symmetric. The safe strategy is:

1. Use y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2
2. Set y(t) equal to the landing height
3. Solve for t (often a quadratic)
4. Use that t in x(t) = x_0 + v_{0x} t

A common error is to use t_{flight} = 2v_{0y}/g even when the landing height differs. That symmetry shortcut is not valid then.

Worked example: standard level-ground range and max height

A projectile is launched from the ground with speed v_0 = 30\ \text{m/s} at angle \theta = 40^\circ. Find (a) time of flight, (b) range, (c) maximum height above launch.

First find components:

v_{0x} = v_0\cos\theta = 30\cos 40^\circ = 23.0\ \text{m/s}

v_{0y} = v_0\sin\theta = 30\sin 40^\circ = 19.3\ \text{m/s}

(a) Time of flight (same launch/landing height):

t_{flight} = \frac{2v_{0y}}{g} = \frac{2(19.3)}{9.8} = 3.94\ \text{s}

(b) Range:

R = v_{0x} t_{flight} = (23.0)(3.94) = 90.6\ \text{m}

(c) Maximum height:

\Delta y_{max} = \frac{v_{0y}^2}{2g} = \frac{(19.3)^2}{2(9.8)} = 19.0\ \text{m}

Notice how the problem never required you to combine x and y into one messy equation; the “independence” structure keeps the work organized.

Worked example: launched from a cliff (quadratic time)

A ball is launched horizontally from a cliff of height h = 45\ \text{m} with speed v_{0x} = 12\ \text{m/s}. Find (a) time to hit the ground and (b) horizontal distance traveled.

Choose origin at launch point, so y_0 = 0. Ground is at y = -45\ \text{m}. Also v_{0y} = 0.

Vertical equation:

y(t) = 0 + 0\cdot t - \frac{1}{2}gt^2 = -\frac{1}{2}gt^2

Set y(t) = -45:

-\frac{1}{2}(9.8)t^2 = -45

t^2 = \frac{90}{9.8} = 9.18

t = 3.03\ \text{s}

Horizontal distance:

x(t) = v_{0x} t = (12)(3.03) = 36.4\ \text{m}

A classic misconception is to think “horizontal launch means it falls straight down at first.” It doesn’t: it begins moving sideways immediately because v_{0x} is nonzero.

Trajectory equation (optional but useful)

Sometimes you’re asked to show the path is parabolic by eliminating t.

From x(t) = x_0 + v_{0x}t (take x_0 = 0 for simplicity):

t = \frac{x}{v_{0x}}

Substitute into vertical equation (take y_0 = 0):

y = v_{0y}\left(\frac{x}{v_{0x}}\right) - \frac{1}{2}g\left(\frac{x}{v_{0x}}\right)^2

This is a quadratic in x, which corresponds to a parabola.

Exam Focus

• Typical question patterns:
  • “Given v_0 and \theta, find time of flight, range, max height, or velocity at a time.”
  • “Projectile launched from height: solve for impact time using y(t), then find horizontal distance.”
  • “At what angle(s) does it land at a given point? (Often requires using both x(t) and y(t) together.)”
• Common mistakes:
  • Using symmetric-flight shortcuts (like t_{flight} = 2v_{0y}/g) when launch and landing heights differ.
  • Setting acceleration equal to -g in both directions, or letting v_x change without air resistance.
  • Forgetting that at the top v_y = 0 but v_x is not.

Relative Motion

Relative motion is the study of how different observers (frames of reference) describe the motion of the same object. In AP Physics C mechanics, you usually stay in the realm of Galilean relativity: speeds are much smaller than the speed of light, so time is shared between observers and velocities add in a straightforward way.

Frames of reference (what changes and what doesn’t)

A frame of reference is essentially an observer’s coordinate system. If you stand on the ground and watch a car go by, you measure one velocity. If you’re in a moving bus and watch that same car, you measure a different velocity.

What changes between frames:

• Measured positions \vec{r}
• Measured velocities \vec{v}

What does not change (in Galilean relativity):

• Time intervals \Delta t (all observers agree on the same time)

This is why relative motion is mostly vector addition and subtraction.

Relative position and velocity relationships

A clean way to avoid sign confusion is to use a consistent notation:

• \vec{r}_{A/B} means “position of A as measured from B”
• \vec{v}_{A/B} means “velocity of A as measured from B”

Then the key vector relationship is:

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

Differentiate with respect to time to get the velocity relation:

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

And differentiate again for acceleration:

\vec{a}_{A/C} = \vec{a}_{A/B} + \vec{a}_{B/C}

In many AP problems, the frames move at constant velocity relative to each other, so \vec{a}_{B/C} = 0 and accelerations are the same in both frames.

Why relative motion matters in 2D kinematics

Relative motion is the bridge between “pure projectile motion” and more realistic situations:

• A plane’s velocity relative to the ground depends on its airspeed and the wind velocity.
• A boat’s velocity relative to shore depends on its velocity relative to the water and the current.
• In projectile problems, you can sometimes simplify by choosing a moving frame (though in AP Mechanics, most problems are solved in the ground frame).

Because velocities are vectors, 2D relative motion is inherently a component problem.

Worked example: boat crossing a river (vector addition)

A boat can move at 4.0\ \text{m/s} relative to the water. The river current is 3.0\ \text{m/s} east relative to the shore. The boat aims due north relative to the water. Find the boat’s velocity relative to the shore.

Define:

• \vec{v}_{B/W}: boat relative to water = 4.0\ \text{m/s} north
• \vec{v}_{W/S}: water relative to shore = 3.0\ \text{m/s} east
• \vec{v}_{B/S}: boat relative to shore (what we want)

Use the relative velocity equation:

\vec{v}_{B/S} = \vec{v}_{B/W} + \vec{v}_{W/S}

Components:

v_{B/S,x} = 3.0\ \text{m/s}

v_{B/S,y} = 4.0\ \text{m/s}

Speed relative to shore:

|\vec{v}_{B/S}| = \sqrt{3.0^2 + 4.0^2} = 5.0\ \text{m/s}

Direction relative to east:

\tan \theta = \frac{4.0}{3.0}

So the boat moves at angle \theta \approx 53^\circ north of east (or equivalently about 37^\circ east of north).

Common mistake: adding speeds 4 + 3 = 7\ \text{m/s} as if the velocities were in the same direction.

Worked example: airplane and wind (solve for heading)

A plane wants to travel due north relative to the ground at 150\ \text{m/s}. A wind blows east at 40\ \text{m/s} relative to the ground. What must the plane’s velocity relative to the air be (magnitude and direction)?

Let:

• \vec{v}_{P/G}: plane relative to ground (desired) = 150\ \text{m/s} north
• \vec{v}_{A/G}: air relative to ground (wind) = 40\ \text{m/s} east
• \vec{v}_{P/A}: plane relative to air (what the plane controls)

Relationship:

\vec{v}_{P/G} = \vec{v}_{P/A} + \vec{v}_{A/G}

So:

\vec{v}_{P/A} = \vec{v}_{P/G} - \vec{v}_{A/G}

Components (take +x east, +y north):

v_{P/A,x} = 0 - 40 = -40\ \text{m/s}

v_{P/A,y} = 150 - 0 = 150\ \text{m/s}

Magnitude:

|\vec{v}_{P/A}| = \sqrt{(-40)^2 + 150^2} = 155\ \text{m/s}

Direction: negative x means “west of north.” The heading angle west of north satisfies:

\tan \phi = \frac{|v_{P/A,x}|}{v_{P/A,y}} = \frac{40}{150}

So \phi \approx 15^\circ west of north.

What this teaches: to cancel an eastward wind and still move due north, the plane must aim slightly into the wind.

Relative motion and projectiles (conceptual connection)

Relative motion thinking helps you understand why projectile motion “splits” into independent axes. If you move to a frame traveling horizontally with the projectile (a frame moving at v_{0x}), then in that moving frame the projectile has no horizontal motion at all and simply goes straight up and down under gravity. That’s not a new physics law—it’s the same motion described from a different inertial frame.

Just be cautious: AP Mechanics typically expects you to stay in one inertial frame (often the ground) unless the problem explicitly invites a frame shift.

Exam Focus

• Typical question patterns:
  • “Boat/plane in wind/current: use \vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C} to find ground speed or required heading.”
  • “One object moving relative to another (e.g., person walking on a train): compute velocity relative to ground.”
  • “Given relative velocity components, find magnitude/direction with vector methods.”
• Common mistakes:
  • Reversing subscripts and accidentally computing the opposite relative velocity (remember \vec{v}_{A/B} = -\vec{v}_{B/A}).
  • Treating relative velocity as scalar addition when directions differ.
  • Mixing coordinate systems mid-problem (changing what counts as +x or +y without updating signs consistently).