When we consider motion we always consider the motion of objects in a reference frame, or coordinate system. An appropriate choice of coordinate system can help enormously in solving a problem.

We need a quantity to express the position of an object. The displacement of an object is a vector which points from the origin of the coordinate system to the object. For now we'll deal with 1 dimensional displacement.

We should also consider the relative displacement of one object from another, or of an object from a position it occupied at an earlier time.

Note that the displacement of an object is not necessarily the same as the distance it has traveled!

If we denote the displacement from a point at time $t_{1}$ as $x_{1}$ and at time $t_{2}$ as $x_{2}$ the change in the displacement in the time interval $\Delta t=t_{2}-t_{1}$ is $\Delta x = x_{2}-x_{1}$.

The average velocity in this time interval is

$\large \bar{v}=\frac{\Delta x}{\Delta t}$

The instantaneous velocity is found by taking the limit as $\Delta t\to 0$

$\large v=^\lim_{\Delta t\to0} \frac{\Delta x}{\Delta t}=\frac{dx}{dt}$

Note that the velocity is the **slope** of the interval, not the length of the arrow!

In the same way as velocity is the rate of change of displacement acceleration is the rate of change of velocity

Average acceleration is

$\large \bar{a}=\frac{\Delta v}{\Delta t}$

The instantaneous acceleration is found by taking the limit as $\Delta t\to 0$

$\large a=^\lim_{\Delta t\to0} \frac{\Delta v}{\Delta t}=\frac{dv}{dt}=\frac{d^2x}{dt^2}$

Using the Labquests and ultrasonic motion detectors we can explore position vs time and velocity vs time graphs. Explore various motion paths by moving an object backwards and forwards in front of the detector and think about the relationship between displacement, velocity and acceleration.

Most smartphones include accelerometers to be able to detect motion and there are apps available which provide access to their readouts.

We can see that the accelerometer measures the acceleration due to gravity. Many smartphones and tablets now add, in addition to the accelerometer, a gyroscope which can determine which direction is down and compensate for the acceleration due to gravity. You can try loading this link on your device and you will get a plot for the acceleration of the three axes of the phone against time (if your device has the necessary sensors).

In classical physics we can easily transform the displacement, velocity and acceleration from one reference frame to another, using Galilean transformations (i.e. just subtract the motion of the reference frame from the motion of the object.)

For example, if I am running at 10km/h (relative to a stationary observer), past someone who is running at 9km/h, my relative velocity to the other person, or, my velocity in their reference frame is 10km/h - 9km/h = 1 km/h.

We should be mindful that all these quantities are vectors, we will later learn how to deal with relative motion in more than one dimension.

Relativity, which we will not be covering in this course, results in a different set of transformations, because in this theory light moves at the same speed in all reference frames. At speeds much less than the speed of light the mathematical limits of relativistic transformations are the same as the Galilean transformations.

These three equations will be very important in this course. Make sure you know how to use them correctly! They apply only when the acceleration $a$ is a constant that does not change with time $t$. In these equations a variable with a subscript 0 refers to the value of that quantity at time $t=0$.

$\large v= v_{0}+at$

$\large x= x_{0}+v_{0}t+\frac{1}{2}at^2$

$\large v^{2}=v_{0}^2+2a(x-x_{0})$

If the acceleration is a constant $a$ which does not change with time then we can say at a time $t$

$\int^{v}_{v_0}\,dv= v(t)-v_0=\int^{t}_0 a\,dt = at$ where $v_{0}$ is the velocity at $t=0$.

so $v= v_{0}+at$

and similarly

$\int^{x}_{x_0}\,dx= x(t)-x_0=\int^{t}_{0} v\,dt = v_{0}t+\frac{1}{2}at^2$

so $x= x_{0}+v_{0}t+\frac{1}{2}at^2$

To find a relation between acceleration and velocity that does not include time

$\large a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$

$\large \int^{x}_{x_{0}}a\,dx=\int^{v}_{v_{0}}v\,dv$

$\large a(x-x_{0})=\frac{1}{2}(v^2-v_{0}^2)$

$\large v^{2}=v_{0}^2+2a(x-x_{0})$

As you can see the equations of motion are simply a result of calculus!

$\large v= v_{0}+at$

$\large x= x_{0}+v_{0}t+\frac{1}{2}at^2$

$\large v^{2}=v_{0}^2+2a(x-x_{0})$

Galileo made a remarkable observation that close to the Earth the acceleration due to gravity is a constant for all objects. Therefore in the absence of air resistance freely falling objects are well described by the equations of motion we just derived.

His observation was especially remarkable given his lack of accurate measurement tools. Today we can do actually do extremely accurate measurements with everyday devices that you may already own!

By taking a video of a freely falling object and processing it so that we can see the position of the object in each frame (which at 30 frames per second is every 0.033 seconds) we can analyze the motion of the object. The tool to do this is here.

As well as using video frames as we did it is also possible to use sound to measure the time taken for objects to fall. This approach is nicely illustrated at Physclips (Section 2.2)

You will need to use the equations of motion!

$v= v_{0}+at$

$x= x_{0}+v_{0}t+\frac{1}{2}at^2$

$v^{2}=v_{0}^2+2a(x-x_{0})$

This is a free fall problem, so you know what the acceleration $a$ is. You can use one of the equations of motion to find out what the velocity is at the top of the window. You can then use another one of the equations to work out the distance required to achieve that velocity when the object falls from rest. The key is to look at the equations for each step and decide which one is the best to help you get from what you know to what you want to know.

Another free fall problem. You know the initial height and velocity and the acceleration and you want to know the height at a given time before the pelican hits the water. To know *where* the pelican is at a given time before it hits the water you're going to have to first work out *when* that is with respect to the initial time.

Break this motion up in to two parts. Make sure to get all your signs right when you take the information you work out for the first part in to the second part. This problem has hints in Mastering Physics to help guide you to the solution.