Basic Kinematics

If you’ve never done physics before, this is a good place to start.

Today we’re gonna cover the most basic aspect of physics. In fact, it’s so simple that most people consider it math. Let’s get started.


What are we interested in mechanics? Most of the time, you want to find the position of an object as a function of time. For now, just consider the one dimensional case. If you set up a convenient zero for your x axis, then you can measure all positions relative to that point.

If an object is to the left, it has a negative position. If it’s to the right, it’s positive, and if it’s exactly on the zero point, than the position is zero. You measure the position in meters, or centimetres, or micrometers, or light years or any length unit you want. And you represent it as \displaystyle x(t).

The t stands for time. Again, you set up a convenient zero point and measure all times relative to that point. For example, if you’re throwing the ball, I could define the zero point in time as the instant the ball goes out of your hand. Then I can assign a positive time for when it hits the ground. Notice I can also assign a negative time. For example, negative two seconds might be when you’re picking the ball up and preparing to throw.

What does x(t) mean? We say x is a function of time. Think of it as a snapshot. At each point in time t, the object of interest is at a specific and unique (no quantum yet!) point x(t).

Ok. So let’s say that x(t) = (5 m/s) \cdot t+3m. This means that at time t, the ball is at the position (5 m/s) \cdot t+3m relative to the origin of the x axis. What else can we say about this function?

It’s a straight line! We say that this object is not accelerating or it is moving at a constant speed. More on this later.

Not all trajectories are straight lines. For example,

x(t) = (5m) \cos{\frac{t}{5s}}.

This is actually the solution to a mass on a spring. I’ll talk more about this in another post.

The next important concept is


You probably already know that velocity equals distance over time. Here’s the problem. If the position of the object is not a straight line, then when you take a finite distance over a corresponding time interval, you are actually taking the average velocity.

 For example, let’s say that I walk at a constant speed of 1 m/s to the grocery store and then turn around to walk back. If the grocery store is 100m from me, it will take me 100s to walk there. Since velocity is distance over time, my average velocity over this interval is just 100m/100s=1m/s exactly as you would expect.

However, let’s take the average velocity over 200s. After this time interval, I made it to the grocery store and I’m back at home, so distance is zero. Therefore velocity is automatically zero!

This example shows that average velocity is not what we are looking for. Now let’s try another experiment. Let’s say that you average over smaller and smaller time intervals. For example, let’s say that your time interval is 0.000001s. Then if you know the average velocity over this interval, then you pretty much know the velocity at this instant. Here’s the math.

\displaystyle v_{avg} = \frac{\Delta x}{\Delta t}

This equation just says that the average velocity is equal to the difference in position over time. Now this is the instantaneous velocity.

\displaystyle v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}

If you’ve done calculus before, you’ll recognize this as a derivative. That is,

\displaystyle  \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}

If you’ve never done calculus, I would recommend Khan Academy calculus videos, just to get a brief overview.

Notice that a derivative is also a function! So now you have a well defined concept of the velocity as a function of time.

\displaystyle v = v(t).

Let’s do a simple problem. If

\displaystyle x(t) = (5 m/s) \cdot t


\displaystyle v(t) = \frac{dx}{dt} = 5 m/s.

Notice that this is a constant function (it does not depend on time), so that explains the term constant speed. Let’s try a harder example.

\displaystyle x(t) = (5m) \cos{\frac{t}{5s}}

Then taking a derivative,

\displaystyle v(t) = \frac{dx}{dt} = (-1 m/s) \cdot \sin{\frac{t}{5s}}

The last concept is


This is the most difficult of the three. Here’s the intuitive explanation. If you move at a constant speed, then your acceleration is zero. But let’s say that you’re gradually moving faster. For example, let’s say that your position is

\displaystyle x(t) = (5 m/s^{s}) \cdot t^{2}

Then the velocity is

\displaystyle v(t) = \frac{dx}{dt} = (10m/s^2) \cdot t

How do you tell if the velocity is increasing? Take another derivative! We define the acceleration to be

\displaystyle a(t) = \frac{dv}{dt} = 10m/s^2

Let’s do another example. If

\displaystyle x(t) = (5 m/s) \cdot t


\displaystyle v(t) = 5 m/s

and so

\displaystyle a(t) = \frac{dv}{dt} = 0.

That explains the term non accelerating. 

We can find another expression for acceleration using a bit of math.

\displaystyle a(t) = \frac{dv}{dt} = \frac{d}{dt}( \frac{dx}{dt}) = \frac{d^{2}t}{dt^2}.

In other words, acceleration is the second derivative of position.

We can find out some interesting things if the acceleration is constant. Let’s call it a. Then we immediately know via integration that

\displaystyle v(t) = at+v_{0},

where v_{0} is the initial velocity at time zero.

If we integrate this again, we get

\displaystyle x(t) = \frac{1}{2}at^2+v_{0}t+x_{0}.

Here x_{0} is the initial position at time zero. If the object is initially still and at the origin, then

\displaystyle x(t) = \frac{1}{2}at^2.


\displaystyle x(t)-x_{0} = \frac{1}{2}at^2+v_{0}t = \frac{2v_0+at}{2} \cdot t

But since v(t) = v_0+at, we can substitute into the above equation to get

\displaystyle x(t)-x_{0} = \frac{v_{0}+(v_{0}+at)}{2} \cdot t = \frac{v_{0}+v(t)}{2} \cdot t


\displaystyle v(t)^2-v_0^2 = (v(t)+v_0)(v(t)-v_0) = (2v_0+at)(at) = 2a(x(t)-x_0)

After substituting into the equation for x(t)-x_0.

In summary, we defined position, velocity, acceleration and derived a bunch of equations for constant acceleration.

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