Motion in a plane with constant acceleration

A puck in a breeze

An air-hockey puck glides across the table to the right, at a steady speed. A fan at the side blows across the table, and pushes the puck up the table with a steady acceleration. What path does it follow?

It does not go in a straight line. It curves, more and more. The trick to finding its path is a surprise: we can treat the two directions as if the other one did not exist.

The equations, as vectors

When the acceleration $\vec{a}$ is constant, the equations of motion from the last chapter still hold. They are now vector equations:

Here $\vec{u}$ is the velocity at $t = 0$, and $\vec{r}_0$ is the starting position.

The same equations, two at a time

Each vector equation is really two ordinary equations, one for each direction:

Along $x$:

Along $y$:

Look closely. The $x$ equations contain only $x$ things, and the $y$ equations only $y$ things. The motion along $x$ and the motion along $y$ do not disturb each other. The only thing they share is the time $t$. This one idea solves every projectile problem in the next course.

Try it. The green arrow is the acceleration, which never changes. The red arrow is the velocity at the time you choose.

What shape is the path?

• If $\vec{a} = \vec{0}$, the velocity never changes. The path is a straight line, at a steady speed.
• If $\vec{a}$ points along $\vec{u}$, or exactly against it, the body speeds up or slows down along one line. The path is a straight line again.
• If $\vec{a}$ points at an angle to $\vec{u}$, the path bends towards $\vec{a}$. It is a parabola.

A worked example

A puck starts at the origin with $\vec{u} = 2\hat{i} + 5\hat{j}\ \text{m/s}$, and has a steady acceleration $\vec{a} = \hat{i} - 2\hat{j}\ \text{m/s}^2$. Find its velocity and position at $t = 3\ \text{s}$.

1. Velocity. $v_x = 2\ \text{m/s} + 1\ \text{m/s}^2 \times 3\ \text{s} = 5\ \text{m/s}$ and $v_y = 5\ \text{m/s} - 2\ \text{m/s}^2 \times 3\ \text{s} = -1\ \text{m/s}$. So $\vec{v} = 5\hat{i} - \hat{j}\ \text{m/s}$, and the speed is $\sqrt{25\ \text{m}^2/\text{s}^2 + 1\ \text{m}^2/\text{s}^2} = 5.10\ \text{m/s}$.
2. Position. $x = 2\ \text{m/s} \times 3\ \text{s} + \tfrac{1}{2} \times 1\ \text{m/s}^2 \times (3\ \text{s})^2 = 10.5\ \text{m}$ and $y = 5\ \text{m/s} \times 3\ \text{s} - \tfrac{1}{2} \times 2\ \text{m/s}^2 \times (3\ \text{s})^2 = 6\ \text{m}$.
3. Distance from the start. $\sqrt{(10.5\ \text{m})^2 + (6\ \text{m})^2} = 12.1\ \text{m}$.
4. One more question. When is the velocity parallel to the $x$ axis? When $v_y = 5 - 2t = 0$, at $t = 2.5\ \text{s}$.

Watch out. You cannot put speeds into $v = u + at$. The speed after $3\ \text{s}$ is not $|\vec{u}| + |\vec{a}| \times 3\ \text{s}$. That would be true only if $\vec{a}$ pointed exactly along $\vec{u}$. Always work part by part, and combine at the end.

Practise with new numbers

Try it in Python

The program works out the puck of the worked example, part by part, and draws its path.

Remember these

• $\vec{v} = \vec{u} + \vec{a}t$ and $\vec{r} = \vec{r}_0 + \vec{u}t + \tfrac{1}{2}\vec{a}t^2$, for a constant $\vec{a}$.
• Split them into $x$ and $y$ equations. The two motions are independent, and share only $t$.
• With $\vec{a}$ at an angle to $\vec{u}$, the path is a parabola.

Check yourself

Try each one on your own first. If you are stuck, open a hint. When you have an answer, check it with Show the answer.

1. A body starts from the origin with $\vec{u} = 3\hat{i}\ \text{m/s}$ and $\vec{a} = 2\hat{j}\ \text{m/s}^2$. Find its velocity and its position after $2\ \text{s}$.

2. A body has a velocity that points north and an acceleration that points east. Which way does its path bend?

3. When is the path of a body with constant acceleration a straight line?