Motion in a Plane with Constant Acceleration

Constant Acceleration in a Plane

When acceleration is constant, the motion in a plane becomes beautifully simple. The key idea is that the two perpendicular components of motion — along the x‑axis and the y‑axis — are completely independent of each other. This means you can treat the motion as two separate one‑dimensional problems, each with constant acceleration, and then combine the results vectorially.

The constant acceleration vector is written as $\vec{a} = a_x \hat{i} + a_y \hat{j}$, where $a_x$ and $a_y$ are constants. The initial velocity at time $t = 0$ is $\vec{v}_0 = v_{0x} \hat{i} + v_{0y} \hat{j}$, and the initial position is $\vec{r}_0 = x_0 \hat{i} + y_0 \hat{j}$.

Velocity as a Function of Time

Since acceleration is constant, the velocity changes linearly with time. For each component, the one‑dimensional formula applies directly:

Writing this as a single vector equation:

This is the first fundamental result. It tells you that the velocity vector at any instant is simply the initial velocity plus the acceleration multiplied by the elapsed time.

Position as a Function of Time

The position vector also follows from the one‑dimensional constant‑acceleration equations applied component‑wise:

In vector form:

This is the second fundamental result. Notice the perfect analogy with the one‑dimensional case — the only difference is that every quantity is now a vector.

A Useful Relation: Velocity and Displacement

There is also a vector relation that connects velocity, displacement, and acceleration without involving time. Start from the scalar product of acceleration with the displacement vector. The displacement is $\Delta \vec{r} = \vec{r} - \vec{r}_0$. Using the position equation:

Now take the dot product of $\vec{a}$ with $\Delta \vec{r}$:

But from the velocity equation, $\vec{v} = \vec{v}_0 + \vec{a} t$, so $\vec{v} - \vec{v}_0 = \vec{a} t$. Taking the dot product of $(\vec{v} + \vec{v}_0)$ with $(\vec{v} - \vec{v}_0)$:

But also $(\vec{v} + \vec{v}_0) \cdot (\vec{a} t) = t (\vec{v} + \vec{v}_0) \cdot \vec{a}$. Since $\vec{v} = \vec{v}_0 + \vec{a} t$, we have $\vec{v} + \vec{v}_0 = 2\vec{v}_0 + \vec{a} t$. Therefore:

Comparing with the expression for $\vec{a} \cdot \Delta \vec{r}$, we see:

Rearranging:

This is the vector analogue of the one‑dimensional equation $v^2 = v_0^2 + 2 a \Delta x$. It is useful when you know the acceleration and displacement but not the time.

The Path of a Particle Under Constant Acceleration

What shape does the trajectory take? To find out, eliminate time between the x and y position equations. Assume for simplicity that $x_0 = y_0 = 0$ and that the acceleration is purely vertical: $\vec{a} = a_y \hat{j}$ (so $a_x = 0$). Then:

$x = v_{0x} t$ …