Motion in Two or Three Dimensions

« Back to AP Physics Guide / Unit 1: Kinematics / Topic 1.5: Motion in Two or Three Dimensions

Visualization of a projectile trajectory broken into independent horizontal and vertical component vectors.

The greatest secret of 2D kinematics is that horizontal and vertical motions are completely blind to each other.

We are finally ready to combine the vector mathematics from Topic 1.1 with the kinematics from Topic 1.2. When an object moves through 2D or 3D space, it has a position vector $\vec{r}(t)$ , a velocity vector $\vec{v}(t)$ , and an acceleration vector $\vec{a}(t)$ .

The golden rule of multi-dimensional physics is the Independence of Motion: what happens in the x-direction has absolutely no effect on what happens in the y-direction. They only share one common variable: Time ( $t$ ).

1. Vector Kinematic Equations

If an object experiences a constant acceleration vector, we can write a single master equation for its position. This is the 3D vector version of the standard kinematic equation you learned in algebra-based physics:

$\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{a}t^2$

Where $\vec{r}_0$ is the initial position vector and $\vec{v}_0$ is the initial velocity vector.

To actually solve problems, you must break this master vector equation down into separate scalar equations for the $\hat{i}$ (x-axis) and $\hat{j}$ (y-axis) directions:

X-Direction: $x(t) = x_0 + v_{0x}t + \frac{1}{2}a_x t^2$
Y-Direction: $y(t) = y_0 + v_{0y}t + \frac{1}{2}a_y t^2$

2. Projectile Motion

Projectile motion is the classic application of 2D kinematics. Once a projectile is launched (and ignoring air resistance), the only force acting on it is gravity. This creates a very specific set of conditions:

Horizontal ( $\hat{i}$ ): There is no force, so $a_x = 0$ .
Therefore, horizontal velocity is constant: $v_x = v_{0x}$ .

Vertical ( $\hat{j}$ ): Gravity pulls downward, so $a_y = -g$ ( $-9.8 \text{ m/s}^2$ ).
Therefore, vertical velocity changes: $v_y(t) = v_{0y} - gt$ .

Concept First: Notice in the simulator how the green horizontal velocity arrow ( $v_x$ ) never changes length! The red vertical arrow ( $v_y$ ) shrinks on the way up, hits zero at the peak, and grows on the way down.

3. Quick AP Practice

📚 Topic 1.5 Mastery Challenge

1. A projectile is launched from the ground with an initial velocity $v_0$ at an angle $\theta$ above the horizontal. Derive an expression for the total time $T$ it remains in the air before hitting the ground.

Check Answer

Time in the air depends only on the y-direction. We know the projectile returns to the ground, so $\Delta y = 0$ .
$\Delta y = v_{0y}t - \frac{1}{2}gt^2$
$0 = (v_0 \sin\theta)T - \frac{1}{2}gT^2$
Factor out $T$ : $0 = T(v_0 \sin\theta - \frac{1}{2}gT)$
Since $T \neq 0$ at landing: $v_0 \sin\theta = \frac{1}{2}gT \Rightarrow \mathbf{T = \frac{2v_0 \sin\theta}{g}}$ .

2. An object’s position vector is given by $\vec{r}(t) = (3t)\hat{i} + (4t - 5t^2)\hat{j}$ . What is the magnitude of the object’s acceleration?