To carry out kinematics calculations, all we need to do is plug the initial conditions into the correct equation of motion and then read out the answer. It is all about the plug-number-into-equation skill. But where do the equations of motion come from? Now that you know a little bit of calculus, you can see how the equations of motion are derived.

Recall the concepts related to the motion of objects (kinematics):

• $t$: the time, measured in seconds [s].
• $x(t)$: the position as a function of time, also known as the equation of motion.
• $v(t)$: the velocity.
• $a(t)$: the acceleration.
• $x_i=x(0), v_i=v(0)$: the initial conditions.

Recall that the purpose of the equations of kinematics is to predict the motion of objects. Suppose that you know the acceleration of the object $a(t)$ at all times $t$. How could you find $x(t)$ starting from $a(t)$?

The equations of motion $x(t)$, $v(t)$ and $a(t)$ are related: \[ a(t) \overset{\frac{d}{dt} }{\longleftarrow} v(t) \overset{\frac{d}{dt} }{\longleftarrow} x(t). \] The velocity is the derivative of the position function and the acceleration is the derivative of the velocity.

If you know the acceleration of an object $a(t)$ as a function of time and its initial velocity $v_i=v(0)$, you can find its velocity $v(t)$ function at all later times. This is because the acceleration function $a(t)$ describes the change in the velocity of the object. If you know that the object started with an initial velocity of $v_i \equiv v(0)$, the velocity at a later time $t=\tau$ is equal to $v_i$ plus the total acceleration of the object between $t=0$ and $t=\tau$: \[ v(\tau)=v_i+\int_0^\tau a(t)\;dt. \]

If you know the initial position $x_i$ and the velocity function $v(t)$ you can find the position function $x(t)$ by using integration again. We find the position at time $t=\tau$ by adding up all the velocity (changes in the position) that occurred between $t=0$ and $t=\tau$: \[ x(\tau) = x_i + \int_0^\tau v(t)\:dt. \]

The procedure for finding $x(t)$ starting from $a(t)$ can be summarized as follows: \[ a(t) \ \ \overset{v_i + \int\!dt}{\longrightarrow} \ \ v(t) \ \ \overset{x_i+ \int\!dt }{\longrightarrow} \ \ x(t). \]

We will now illustrate how to apply this procedure for the important special case of motion with constant acceleration.

Consider an object undergoing uniform acceleration motion (UAM) with acceleration function $a(t) =a$. Suppose that we know the initial velocity of $v_i \equiv v(0)$, and you want to find the velocity at later time $t=\tau$. We have to compute the following integral: \[ v(\tau) =v_i+ \int_0^\tau a(t)\;dt = v_i + \int_0^\tau a \ dt = v_i + a\tau. \] The velocity as a function of time is given by the initial velocity $v_i$ plus the integral of the acceleration.

If you know the initial position $x_i$ and the velocity function $v(t)$ you can find the position function $x(t)$ by using integration again. The formula is \[ x(\tau) = x_i + \int_0^\tau v(t)\:dt = x_i + \int_0^\tau (at+v_i) \; dt = x_i + \frac{1}{2}a\tau^2 + v_i\tau. \]

Note that the above calculations required the knowledge of the initial conditions $x_i$ and $v_i$. This is because the integral calculations tell us the change in the quantities relative to their initial values.

The fourth equation of motion \[ v_f^2 = v_i^2 + 2a(x_f-x_i) \] can be derived by combining the equations of motion $v(t)$ and $x(t)$.

Consider squaring both sides of the velocity equation $v_f = v_i + at$ to obtain \[ v_f^2 = (v_i + at)^2 = v_i^2 + 2av_it + a^2t^2 = v_i^2 + 2a(v_it + \frac{1}{2}at^2). \] We can recognize the term in the bracket is equal to $\Delta x = x(t)-x_i=x_f-x_i$.

According to Newton's second law of motion, forces are the cause of acceleration and the formula that governs this relationship is \[ F_{net}=ma, \] where $F_{net}$ is the magnitude of the net force on the object.

In a later chapter, we will learn about dynamics which is the study of the different kinds of forces that can act on objects: the gravitational force $\vec{F}_g$, the spring force $\vec{F}_s$, the friction force $\vec{F}_f$, the electric force $\vec{F}_e$, the magnetic force $\vec{F}_b$ and many others. To find the acceleration on an object we must add together (as vectors) all of the forces which are acting on the object and divide by the mass \[ \sum \vec{F} = F_{net}, \qquad \Rightarrow \qquad a = \frac{1}{m} F_{net}. \]

The physics procedure for predicting the motion of objects can be summarized as follows: \[ \frac{1}{m} \underbrace{ \left( \sum \vec{F} = \vec{F}_{net} \right) }_{\text{dynamics}} = \underbrace{ a(t) \ \overset{v_i+ \int\!dt }{\longrightarrow} \ v(t) \ \overset{x_i+ \int\!dt }{\longrightarrow} \ x(t) }_{\text{kinematics}}. \]

The force of gravity on a object of mass $m$ on the surface of the earth is given by $\vec{F}_g=-mg\hat{y}$, where $g=9.81$[m/s$^2$] is the gravitational constant. Recall that we said that an object is in free fall when the only force acting on it is the force of gravity. In this case, Newton's second law tells us that \[ \begin{align*} \vec{F}_{net} &= m\vec{a} \nl -mg\hat{y} &= m\vec{a}. \end{align*} \] Dividing both sides by the mass we see that the acceleration of an object in free fall is $\vec{a} = -9.81\hat{y}$.

It is interesting to note that the mass of the object does not affect its acceleration during free fall. The force of gravity is proportional to the mass of the object, but the acceleration is inversely proportional to the mass of the object so overall we get $a_y = -g$ for objects in free fall regardless of their mass. This observation was first made by Galileo in his famous Leaning Tower of Pisa experiment during which he dropped a wooden ball and a metal ball (of same shape but different mass) from the Leaning Tower of Pisa and observed that they fall to the ground in the same time. Search for “Apollo 15 feather and hammer drop” on YouTube to see a similar experiment performed on The Moon.

You may have noticed that in the last couple of paragraphs we started putting little arrows on top of certain quantities. The arrow is used to remind you that forces are vector quantities. Before we proceed with the physics lessons, we must make a mathematical digression and introduce vectors.