Motion in two dimensions

Topics 1.1 through 1.4 treated motion mostly as 1D: one axis, one velocity component, one acceleration. Topic 1.5 stretches kinematics into the plane. Position, velocity, and acceleration are now vector functions of time, and calculus acts on them.

The vector form is direct. With Cartesian unit vectors $\hat{\text{ı}}$ and $\hat{\text{ȷ}}$:

$\vec{r}(t) = x(t)\,\hat{\text{ı}} + y(t)\,\hat{\text{ȷ}}, \qquad \vec{v}(t) = \dfrac{d\vec{r}}{dt}, \qquad \vec{a}(t) = \dfrac{d\vec{v}}{dt}.$

The headline idea is the independence of perpendicular axes: the $\hat{\text{ı}}$-motion is set entirely by the $\hat{\text{ı}}$-components of the forces on the object, and the $\hat{\text{ȷ}}$-motion entirely by the $\hat{\text{ȷ}}$-components. For projectile motion under uniform gravity with no air resistance, the two axes share only the time $t$ and nothing else crosses. In more general problems a velocity-dependent force like drag, or a constraint linking the directions, can couple the components, so check the force law before splitting the axes.

For a projectile near Earth's surface, the only force is gravity, which acts along $-\hat{\text{ȷ}}$. The $\hat{\text{ı}}$-component of velocity is therefore constant throughout the flight; the $\hat{\text{ȷ}}$-component changes at rate $-g$. The two motions look very different (uniform along $\hat{\text{ı}}$, uniformly accelerated along $\hat{\text{ȷ}}$), but they happen in the same body at the same time. The curved trajectory $y(x)$ is what an outside observer sees; the underlying physics is two simple 1D motions glued together at the shared $t$.

The calculus side adds one more move. Differentiating a vector function means differentiating each scalar component and carrying the unit vectors through unchanged (they are constants in an inertial frame). Integrating works the same way, with one initial-condition constant per component. The trap is to forget half the work: differentiate only one component, drop the other, or treat a unit vector as if it had a non-zero derivative.

The diagnostic targets three mistakes: letting one axis affect the other (it doesn't), dropping a component or writing a phantom unit-vector derivative when you differentiate, and mixing up speed (a scalar) with velocity (a vector).

A point particle has a position vector from the origin of a chosen inertial frame:

$\vec{r}(t) = x(t)\,\hat{\text{ı}} + y(t)\,\hat{\text{ȷ}}.$

In three dimensions we add a $z(t)\,\hat{k}$ term. The AP exam only tests 2D quantitatively, so we'll borrow $\hat{k}$ in notation when it helps but keep every worked problem in the plane.

Velocity is the time derivative of $\vec{r}$. Because $\hat{\text{ı}}$ and $\hat{\text{ȷ}}$ are constants in an inertial frame, their derivatives are zero, and differentiation acts only on the scalar components:

$\vec{v}(t) = \dfrac{d\vec{r}}{dt} = \dfrac{dx}{dt}\,\hat{\text{ı}} + \dfrac{dy}{dt}\,\hat{\text{ȷ}} = v_x(t)\,\hat{\text{ı}} + v_y(t)\,\hat{\text{ȷ}}.$

Acceleration is the time derivative of $\vec{v}$, again component-by-component:

$\vec{a}(t) = \dfrac{d\vec{v}}{dt} = \dfrac{d^2 x}{dt^2}\,\hat{\text{ı}} + \dfrac{d^2 y}{dt^2}\,\hat{\text{ȷ}}.$

• Differentiation is component-wise. $\dfrac{d}{dt}\big[f(t)\,\hat{\text{ı}} + g(t)\,\hat{\text{ȷ}}\big] = f'(t)\,\hat{\text{ı}} + g'(t)\,\hat{\text{ȷ}}$. The unit vectors carry through unchanged; the scalars get differentiated.
• Integration is component-wise. $\int \vec{a}(t)\,dt = \Big[\int a_x(t)\,dt\Big]\,\hat{\text{ı}} + \Big[\int a_y(t)\,dt\Big]\,\hat{\text{ȷ}}$. Each component picks up its own constant of integration, set by its own initial condition.
• Unit vectors are constants. $d\hat{\text{ı}}/dt = 0$ and $d\hat{\text{ȷ}}/dt = 0$ in any inertial frame. Product-rule terms that involve derivatives of unit vectors vanish; do not write them.
• Speed is the magnitude. $|\vec{v}(t)| = \sqrt{v_x^2 + v_y^2}$. Speed is a scalar; it loses the directional information that the vector carries.

These four rules carry through every projectile problem and every kinematics integral for the rest of the course. Everything that follows is an application.

Position, velocity, and acceleration are vector functions of time, related by differentiation: $\vec{v} = d\vec{r}/dt$ and $\vec{a} = d\vec{v}/dt$. The unit vectors are constants in any inertial frame, so calculus acts on each scalar component separately. Integration reverses differentiation, with one constant per component.

Newton's second law in two dimensions reads $\vec{F} = m\vec{a}$, which is one vector equation. Read in components, it is two independent scalar equations:

$F_x = m a_x, \qquad F_y = m a_y.$

The $\hat{\text{ı}}$-component of acceleration is set by the $\hat{\text{ı}}$-component of the net force. The $\hat{\text{ȷ}}$-component is set by the $\hat{\text{ȷ}}$-component of the net force. The two axes are governed by separate equations. They share only the mass $m$ and the time $t$.

The consequence: if no force has an $\hat{\text{ı}}$-component, then $a_x = 0$ for all $t$, so $v_x$ is constant for all $t$. Nothing the $\hat{\text{ȷ}}$-motion does can change the $\hat{\text{ı}}$-motion. The two motions happen in the same particle at the same time, but they obey different equations.

For a projectile near the Earth's surface, the only force is $\vec{F}_{grav} = -mg\,\hat{\text{ȷ}}$. Reading components: $F_x = 0$ and $F_y = -mg$. So $a_x = 0$ and $a_y = -g$. The $\hat{\text{ı}}$-motion is uniform (constant velocity); the $\hat{\text{ȷ}}$-motion is uniformly accelerated at $-g$. Same start, same end, but otherwise they do their own thing.

The classic intuition trap is "the ball is falling, so it must be slowing down horizontally." This couples the axes through a mechanism that isn't there. No force has an $\hat{\text{ı}}$-component, so $v_x$ does not change. The $\hat{\text{ȷ}}$-motion has no say in what $v_x$ does.

The same logic gives the famous result that a bullet fired horizontally and a bullet dropped from the same height land at the same time (assuming flat ground and no air resistance). Both have $v_{0y} = 0$, both have $a_y = -g$, so both satisfy $h = \tfrac{1}{2}gt^2$ with the same $t$. The fired bullet's horizontal velocity has no $\hat{\text{ȷ}}$-component and cannot affect the time of fall. The fired bullet lands far downrange; the dropped one lands directly below. They land at the same instant.

Perpendicular axes are independent: each axis has its own $F = ma$, with no coupling between them. In projectile motion near Earth, gravity has no $\hat{\text{ı}}$-component, so $v_x$ is constant for the whole flight. The horizontal and vertical motions share only $t$.

A projectile is launched from level ground at $v_0 = 20$ m/s at $\theta = 30^\circ$ above the horizontal. Take $\hat{\text{ı}}$ along the launch direction's horizontal component, $\hat{\text{ȷ}}$ upward, $g = 10$ m/s$^2$. Air resistance negligible.

Decompose the launch velocity into components.

$v_{0x} = v_0\cos\theta = 20\cos 30^\circ \approx 17.3 \text{ m/s}, \qquad v_{0y} = v_0\sin\theta = 20\sin 30^\circ = 10 \text{ m/s}.$

So $\vec{v}_0 = 17.3\,\hat{\text{ı}} + 10\,\hat{\text{ȷ}}$ m/s. The acceleration during the flight is $\vec{a} = -10\,\hat{\text{ȷ}}$ m/s$^2$; nothing pushes along $\hat{\text{ı}}$.

Write the component equations of motion. Integrate $\vec{a}$ twice to get $\vec{v}(t)$ and $\vec{r}(t)$:

$v_x(t) = 17.3 \text{ m/s}, \qquad v_y(t) = 10 - 10 t \text{ m/s}.$

$x(t) = 17.3\, t \text{ m}, \qquad y(t) = 10 t - 5 t^2 \text{ m}.$

The $\hat{\text{ı}}$-velocity is constant at $17.3$ m/s for the entire flight; the $\hat{\text{ȷ}}$-velocity decreases linearly. Each component carries its own initial condition: $v_x(0) = 17.3$, $v_y(0) = 10$, $x(0) = 0$, $y(0) = 0$.

Find the time of flight. The ball returns to launch height when $y(t) = 0$, which gives $t(10 - 5t) = 0$, so $t = 0$ (launch) or $t = 2$ s (landing). Time of flight: $t_f = 2$ s.

Equivalently, find when $v_y = 0$: $10 - 10t = 0$ gives $t = 1$ s at the apex. The motion is symmetric in time around the apex (since $a_y$ is constant), so $t_f = 2 t_{apex} = 2$ s.

Find the range. Horizontal velocity is constant, so $R = v_x \cdot t_f = 17.3 \times 2 \approx 34.6$ m. The range comes entirely from the constant $\hat{\text{ı}}$-velocity; the $\hat{\text{ȷ}}$-motion only sets how long the flight lasts.

Find the maximum height. Maximum height comes from the $\hat{\text{ȷ}}$-motion at $t_{apex}$:

$y_{max} = y(1) = 10(1) - 5(1)^2 = 5 \text{ m}.$

Or symbolically: $y_{max} = v_{0y}^2 / (2g) = 100/20 = 5$ m. The horizontal velocity does not appear in the height formula because it has no $\hat{\text{ȷ}}$-component.

Find the velocity at the apex. At $t = 1$ s, $v_x = 17.3$ m/s and $v_y = 0$. So $\vec{v}(1) = 17.3\,\hat{\text{ı}}$ m/s, with speed $|\vec{v}(1)| = 17.3$ m/s. The ball is not at rest at the apex; only its $\hat{\text{ȷ}}$-velocity is zero. Horizontally, it still moves at $17.3$ m/s.

Find the velocity at landing. At $t = 2$ s, $v_x = 17.3$ m/s and $v_y = 10 - 20 = -10$ m/s. So $\vec{v}(2) = 17.3\,\hat{\text{ı}} - 10\,\hat{\text{ȷ}}$ m/s, with speed $|\vec{v}(2)| = \sqrt{300 + 100} = \sqrt{400} = 20$ m/s. Same speed as at launch, by energy conservation (same height, no air resistance), with the $\hat{\text{ȷ}}$-component reversed in sign.

Every quantity in this problem split into two independent pieces. The $\hat{\text{ı}}$-motion is uniform; the $\hat{\text{ȷ}}$-motion is uniformly accelerated. The two motions share only $t$. The curved trajectory is what an outside observer sees after both motions play out simultaneously.

Projectile motion is two uncoupled 1D problems sharing only the time variable. Decompose the launch velocity into $v_{0x}$ and $v_{0y}$, write a constant-velocity equation for $x(t)$ and a constant-acceleration equation for $y(t)$, and read off whatever the problem asks. The horizontal velocity stays at $v_{0x}$ for the entire flight.

Projectile motion is the constant-acceleration special case. When $\vec{a}$ varies with time (or position, or velocity), the kinematic equations don't apply and we go back to calculus. The component-wise structure stays the same.

Given a position function $\vec{r}(t) = x(t)\,\hat{\text{ı}} + y(t)\,\hat{\text{ȷ}}$, the velocity is:

$\vec{v}(t) = \dfrac{d\vec{r}}{dt} = x'(t)\,\hat{\text{ı}} + y'(t)\,\hat{\text{ȷ}}.$

And the acceleration:

$\vec{a}(t) = \dfrac{d\vec{v}}{dt} = x''(t)\,\hat{\text{ı}} + y''(t)\,\hat{\text{ȷ}}.$

Each scalar component is differentiated using single-variable calculus. The unit vectors are constants and carry through untouched.

Worked example, differentiation. A particle has $\vec{r}(t) = (4t - t^2)\,\hat{\text{ı}} + 2t^3\,\hat{\text{ȷ}}$, in meters and seconds. Find $\vec{v}(t)$ and $\vec{a}(t)$. Differentiating:

$\vec{v}(t) = (4 - 2t)\,\hat{\text{ı}} + 6t^2\,\hat{\text{ȷ}} \text{ m/s}, \qquad \vec{a}(t) = -2\,\hat{\text{ı}} + 12 t\,\hat{\text{ȷ}} \text{ m/s}^2.$

Both components are time-varying; both have to be carried through every step. Dropping the $\hat{\text{ȷ}}$-component on either derivative would lose half the answer.

Worked example, integration. A particle has acceleration $\vec{a}(t) = 4\,\hat{\text{ı}} + 6 t\,\hat{\text{ȷ}}$ m/s$^2$, with $\vec{v}(0) = 0$ and $\vec{r}(0) = 0$. Integrating $\vec{a}$ component-by-component:

$v_x(t) = \int_0^t 4\,d\tau = 4 t, \qquad v_y(t) = \int_0^t 6\tau\,d\tau = 3 t^2.$

So $\vec{v}(t) = 4 t\,\hat{\text{ı}} + 3 t^2\,\hat{\text{ȷ}}$ m/s. Integrate again to get position:

$x(t) = \int_0^t 4\tau\,d\tau = 2 t^2, \qquad y(t) = \int_0^t 3\tau^2\,d\tau = t^3.$

So $\vec{r}(t) = 2 t^2\,\hat{\text{ı}} + t^3\,\hat{\text{ȷ}}$ m. Each component gets its own antiderivative and its own initial-condition match. The constant of integration is set component by component; equivalently, those component constants collect into a single initial-value vector.

The general statements are $\vec{v}(t) = \vec{v}_0 + \int_0^t \vec{a}(\tau)\,d\tau$ and $\vec{r}(t) = \vec{r}_0 + \int_0^t \vec{v}(\tau)\,d\tau$. The initial vectors $\vec{v}_0$ and $\vec{r}_0$ are the constants of integration, one component at a time.

Differentiation and integration of vector functions are component-wise operations. Treat $\hat{\text{ı}}$ and $\hat{\text{ȷ}}$ as constants and act on each scalar component separately, matching each component's initial condition. Forgetting one component is the most common slip.

The component functions $x(t)$ and $y(t)$ describe the motion in time. A useful alternative is to eliminate $t$ and write $y$ as a function of $x$, giving the shape of the trajectory without the timing.

For projectile motion launched from the origin at speed $v_0$ at angle $\theta$:

$x(t) = (v_0 \cos\theta)\,t, \qquad y(t) = (v_0 \sin\theta)\,t - \tfrac{1}{2} g t^2.$

Solve the first equation for $t = x / (v_0 \cos\theta)$ and substitute into the second:

$y(x) = x \tan\theta - \dfrac{g\, x^2}{2 v_0^2 \cos^2\theta}.$

This is a parabola opening downward; the trajectory of a projectile launched from level ground with a nonzero horizontal velocity, under uniform gravity with negligible air resistance, is a downward-opening parabola. The first term ($x \tan\theta$) is the line the projectile would follow with no gravity; the second term is the gravitational sag. The two together produce the parabolic arc.

Apex, range, and time-of-flight follow as derived quantities. The range on level ground is found by setting $y(x) = 0$ and solving the non-trivial root:

$R = \dfrac{v_0^2 \sin(2\theta)}{g}.$

For our $v_0 = 20$ m/s, $\theta = 30^\circ$, $g = 10$ m/s$^2$ example, $\sin 60^\circ = \sqrt{3}/2 \approx 0.866$, so $R = 400 \times 0.866 / 10 \approx 34.6$ m. Matches the time-based calculation in §4.

The maximum height sits at $x = R/2$ by symmetry:

$y_{max} = \dfrac{v_0^2 \sin^2\theta}{2 g} = \dfrac{400 \times 0.25}{20} = 5 \text{ m}.$

And the time of flight on level ground:

$t_f = \dfrac{2 v_0 \sin\theta}{g} = \dfrac{20}{10} = 2 \text{ s}.$

The formulas above all come from the same two component equations; we just solve them for different unknowns. The point of this section is not the formulas (the AP equation sheet does not list range or max height) but the principle: treat the two components independently, then combine the answers at the end.

Eliminating $t$ between $x(t)$ and $y(t)$ gives the trajectory shape $y(x)$, a downward-opening parabola for projectile motion. Apex, range, and time-of-flight are derived quantities, obtained from the same two independent component equations.

All three below come from mishandling the two axes — coupling them when they're independent, dropping one when you differentiate, or collapsing both into a single speed.