Position, displacement, velocity and acceleration in a plane

A car that does not go straight

A toy car runs across a playground in a curve. At every moment we want three answers. Where is it? How fast is it going? Which way?

In the last chapter, a body moved along one line. A plus or minus sign was enough to give its direction. On a playground that is not enough. Every quantity of motion becomes a vector, with an $x$ part and a $y$ part. The good news: each part behaves exactly like the motion along a line that you already know.

Position and displacement

Choose an origin O and two axes. A body at the point $(x, y)$ has the position vector

Say the car is at P at the time $t_1$, and at Q at a later time $t_2$. Its displacement is the change in position:

It is the straight arrow from P to Q. The curve the car took in between does not matter.

Average velocity

The average velocity is the displacement divided by the time taken:

It points along the displacement, from P straight towards Q.

Watch out. Average speed and average velocity are not the same on a curve. A runner goes half way round a circular track of radius $35\ \text{m}$ in $22\ \text{s}$. She runs $\pi \times 35\ \text{m} = 110\ \text{m}$, so her average speed is $5.0\ \text{m/s}$. But her displacement is the diameter, $70\ \text{m}$, so her average velocity is only $3.2\ \text{m/s}$.

Velocity at an instant

Now let Q slide closer and closer to P. The arrow from P to Q gets shorter, and it turns. When Q is almost on top of P, the arrow lies along the tangent to the path at P. The average velocity has become the velocity at that instant:

See it happen. The car's path here is $x = 4t$ and $y = 5t - t^2$.

So, velocity always points along the tangent to the path, in the direction of motion. Its size is the speed, $v = \sqrt{v_x^2 + v_y^2}$, and its angle with the $x$ axis is given by $\tan\theta = \dfrac{v_y}{v_x}$.

Acceleration

The acceleration is the rate of change of velocity:

Think first. A car goes round a bend at a steady $40\ \text{km/h}$. Is it accelerating?

Yes. Its speed does not change, but its direction does, and velocity includes direction. Any change in velocity, in size or in direction or both, is an acceleration. This is the big new idea of motion in a plane. We meet it again in uniform circular motion.

Also notice: the acceleration does not have to point along the velocity. For a thrown ball, the velocity points along the path, but the acceleration always points straight down.

A worked example

The toy car moves so that $x = 4t$ and $y = 5t - t^2$, in metres, with $t$ in seconds. Find its velocity and acceleration at $t = 1\ \text{s}$.

1. Velocity parts. $v_x = \dfrac{d}{dt}(4t) = 4\ \text{m/s}$ and $v_y = \dfrac{d}{dt}(5t - t^2) = 5 - 2t = 3\ \text{m/s}$ at $t = 1\ \text{s}$.
2. Speed and direction. $v = \sqrt{(4\ \text{m/s})^2 + (3\ \text{m/s})^2} = 5\ \text{m/s}$, at $\tan^{-1}\dfrac{3\ \text{m/s}}{4\ \text{m/s}} = 36.9^\circ$ above the $x$ axis.
3. Acceleration. $a_x = \dfrac{d}{dt}(4) = 0$ and $a_y = \dfrac{d}{dt}(5 - 2t) = -2\ \text{m/s}^2$. So $\vec{a} = -2\hat{j}\ \text{m/s}^2$, the same at every moment.
4. A bonus. When is the velocity horizontal? When $v_y = 5 - 2t = 0$, at $t = 2.5\ \text{s}$. That is the highest point of the path.

Practise with new numbers

Try it in Python

The program finds the car's average velocity over shorter and shorter times after $t = 1\ \text{s}$, and compares it with the exact velocity.

Remember these

• $\vec{r} = x\hat{i} + y\hat{j}$ and $\Delta\vec{r} = \vec{r}_2 - \vec{r}_1$.
• $\vec{v}_{\text{av}} = \Delta\vec{r}/\Delta t$ points along the displacement. $\vec{v} = d\vec{r}/dt$ points along the tangent.
• $\vec{a} = d\vec{v}/dt$. A change in the size or the direction of the velocity is an acceleration.

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 moves from $(2, -1)$ to $(8, 7)$, in metres, in $4\ \text{s}$. Find its average velocity and its size.

2. A body moves so that $x = 3t$ and $y = 4t$, in metres. Find its speed and its acceleration.

3. In which direction does the velocity point, for a body on a curved path?

4. A cyclist rides round a circular track at a steady speed. Is the acceleration zero?