Equations Of Motion

The equations of motion of kinematics describe the most fundamental concepts of motion of an object. These equations govern the motion of an object in 1D, 2D and 3D. They can easily be used to calculate expressions such as the position, velocity, or acceleration of an object at various times.

Do you know the speed of the world fastest human? It's a mind blowing \(12.4 \text{ m/s}!\) But what does this speed signify? In this wiki, the three equations of motion in 1D kinematics will be derived so that quantities like the speed of a runner may be predicted. Note that this wiki is not to be confused with Newton's laws of motion; though Newton's laws also describe motion, they apply to the motion of systems with applied forces.

World's fastest human, Usain Bolt in full swing[1]

Distance and Displacement:

If an object changes its position with respect to time, the measure of the change in the position of the object can be taken in two different ways. One can consider the actual measure of the path traversed or the shortest measure between the two points.

Difference between distance and displacement[2]

This is where the difference between two very fundamental concepts of physics comes. The first way to measure the change in position of the object is called distance, while the second is called displacement.

The actual measure of the total change in the position of an object (in a certain period of time) is called distance. It is a scalar quantity and thus gives the magnitude only.

The shortest measure of the net change in the position of an object (in a certain time interval) is called displacement. It is a vector quantity and thus gives both a magnitude and the direction.

\[\] Speed and Velocity:

What happens if the distance/displacement of an object changes with respect to time? This is where the idea of speed and velocity come in.

The distance traveled by an object per unit time is called speed. It is a scalar quantity and thus gives a magnitude only: \[\text{speed} =\dfrac{\text{distance}} {\text{time}}.\]

The rate of change of displacement of an object with respect to time, or the rate of change of position, is called velocity. Graphically, it is the slope of the displacement function. It is a vector quantity and thus gives both a magnitude and direction: \[\text{velocity} =\dfrac{\text{displacement}} {\text{time}}=\dfrac{\Delta x}{\Delta t}.\]

Below is the graphical representation of the velocity function:

Graph of an object starting at \(S_1\)

Acceleration:

If velocity is the rate of change of displacement, what is the rate of change of velocity? This is the acceleration.

The change in velocity of an object per unit time, or the rate of change of velocity function with respect to time, is defined as the acceleration. Graphically, it is the slope of the velocity function. Acceleration is a vector quantity and thus gives both a magnitude and direction: \[\text{acceleration}=\dfrac{\text{change in velocity}} {\text{time}}=\dfrac{\Delta v}{\Delta t}.\]

This means that acceleration is the derivative of the velocity function. Thus, it is also the second derivative of the displacement function.

\[a(t)=\dfrac{d}{dt}v=\dfrac{d^2}{dt^2}x.\]

We can verify this by the following graph which shows the slope of the acceleration function as the derivative of the velocity function:

Graph of an object with an initial velocity u

Graphs are very interactive and help summarize various things in a single place. They are consequently useful in identifying, tracking, and modelling the motion of an object. The following types of graphs may be useful in kinematics: position-time graphs and velocity-time graphs.

Position-time Graphs:

Position-time graphs are the most basic graphs in kinematics. They allow description of the motion of an object with respect to both position and time. The \(y\)-axis of these graphs represent the displacement of the object, and the \(x\)-axis represents the time. Thus, the slope of a position-time graph gives the velocity of the object.

Find the velocity of the particle over the interval \((1,2)\).

---

One can see that the slope of the position-time graph gives us the velocity over the time period. Thus,

\[\begin{align} v&=\dfrac{s_2-s_1}{t_2-t_1}\\ &=\dfrac{10-7.5}{2-1}\\ &=2.5\text{ m/s}. \end{align}\]

Also, one can see that the displacement of this object is given by the function \(s(t)=2.5t+5\). Thus, one can find the velocity of this function which is nothing but its derivative:

\[v(t)=\dfrac{d}{dt}(2.5t+5)=2.5\text{ m/s}.\]

Velocity-time Graphs:

These graphs allow computation of the velocity of the object at a given instant of time. The \(y\)-axis represents the velocity while the \(x\)-axis represents the time. They can help to calculate two things mainly:

1. displacement of the object over an interval,

2. acceleration of the object over an interval.

From the graph above one can observe that the slope of the velocity-time graph gives the acceleration:

\[(\text{Slope of a velocity time graph})=a=\dfrac{v_2-v_1}{t_2-t_1}.\]

Also, one can observe that the area under the velocity time graph (which is nothing but the integral) gives the displacement over the interval:

\[(\text{Area under the velocity time graph})=s=\displaystyle\int_{t_1}^{t_2} v(t).\]

Find the acceleration and the displacement traversed by the object over the interval \((2,4)\).

---

The slope of the velocity-time graph gives the acceleration over the time period, thus

\[\begin{align} a&=\dfrac{v_2-v_1}{t_2-t_1}\\ &=\dfrac{10-6}{4-2}\\ &=2\text{ m/s}^2. \end{align}\]

The area under the graph gives the displacement, which is the area of the trapezoid:

\[\begin{align} s&=\dfrac{h(a+b)}2\\ &=\dfrac{2(6+10)}{2}\\ &=16\text{ m}. \end{align}\]

Also, one can deduce that the graph is given by the function: \(v(t)=2t+2\). Thus the acceleration would be the derivative

\[\begin{align} a&=\dfrac{d}{dt}(2t+2)\\ &=2\text{ m/s}^2. \end{align}\]

The displacement is the integral

\[\begin{align} s&=\displaystyle\int_{2}^{4}2t+2 dt\\ &=\left[t^2+2t\right]_{2}^{4}\\ &=4^2+8-\left(2^2+4\right)\\ &=16\text{ m}. \end{align}\]

Though using graphs may appear messier than using simple equations to solve the questions, it can provide important intuition in solving physical problems and give clues as to how to derive equations of motion in complex cases. In the following sections, the velocity-time graphs are used to obtain the fundamental equations which govern the motion of objects.

Consider an object moving so that its velocity-time graph is a straight line. Such an object is undergoing constant acceleration, since the slope of the graph is constant. The first equation of motion gives the final velocity after a time \(t\) for these objects, given an initial velocity \(v_0\).

\[v=v_0+a \Delta t\]

The graph of the motion of the object

Suppose the object is observed from \(t_1\) to \(t_2\). Assume that the object was moving with velocity \(v_0\) at \(t_1\), and it is moving with a velocity of \(v\) at \(t_2\). Then this equation of motion can be proved graphically:

From the definition of acceleration, the slope of the velocity-time graph gives us the acceleration, so

\[\begin{align} a=\dfrac{(\text{change in velocity})}{(\text{time period})}&=\dfrac{v-v_0}{t_2-t_1}\\ &=\dfrac{v-v_0}{\Delta t}\\ \Rightarrow v &= v_0 + a\Delta t . \end{align}\]

An object starts from rest, and reaches a velocity of \(25\text{ m/s}\) in \(4\) seconds. Find the acceleration of the object.

---

The problem gives that \(v_0=0\text{ m/s}\), \(v=25\text{ m/s}\), and \(t=4\). Substituting these values in the first equation of motion, one obtains

\[\begin{align} v&=v_0+at\\ 25&=0+a(4)\\ \Rightarrow a&=\dfrac{25}{4}=6.25 ~\left(\si[per-mode=symbol]{\meter\per\second\squared}\right). \end{align}\]

An object is thrown vertically upwards with an initial velocity of \(20\text{ m/s}\). Find the time taken by it to reach the maximum height.

---

When the object reaches its maximum height, the velocity is \(0\text{ m/s}\). Since the object is moving against the force of gravity, its acceleration is \(-g\approx -10\text{ m/s}^2\). Thus, from the first equation of motion, one obtains

\[\begin{align} t&=\dfrac{(v-u)}{-g}\\ &=\dfrac{(0-20)}{-10}\\ &=2\text{ (sec)}. \end{align}\]

The second equation of motion gives the displacement of an object under constant acceleration:

\[x = x_0 + v_0 t + \frac{1}{2}at^{2}.\]

The graph of the motion of the object

Suppose an object is observed from \(t_1\) to \(t_2\). Assume that the object was moving with velocity \(v_0\) at \(t_1\), and it was moving with a velocity of \(v\) at \(t_2\). Again, the equation can be proved graphically:

The (signed) area under the velocity-time graph gives the displacement of the object over the interval. In this case, the area under the function between \(t_1\) and \(t_2\) is the area of upper triangle plus the area of the lower rectangle:

The area of the triangle is

\[\begin{align} \text{Area(triangle)}&=\dfrac 12\times\text{base}\times\text{height}\\ &=\dfrac 12\times(t_2-t_1)\times(v-v_0)\\ &= t\times a t \qquad (\text{since } v-v_0=at)\\ &=a t^2. \end{align}\]

The area of the rectangle is

\[\begin{align} \text{Area(rectangle)}&=\text{length}\times\text{width}\\ &= (t_2-t_1)\times v_0\\ &=v_0 t. \end{align}\]

Thus, the total area is

\[x=v_0 t+ \dfrac 12 a t^2.\]

One can also directly integrate both sides of the first equation of motion to obtain

\[\begin{align} \displaystyle\int_{t_1}^{t_2}v(t) \ dt &=\displaystyle\int_{t_1}^{t_2}v_0+a t \ dt\\ &=v_0 \Delta t+\dfrac 12 a\Delta t^2. \end{align}\]

Since this gives the displacement, to find the position \(x\), the initial position \(x_0\) must be included as an additive constant.

Corollary of The Second Equation of Motion:
The second equation of motion also has a corollary, which gives the distance traveled by a body in the \(n^\text{th}\) second of motion. This is \[s_{n^\text{th}}=v_0+a\left(n-\frac{1}{2}\right).\]

The corollary is proved via the second equation of motion. Since

\[x=v_0 \Delta t + \frac{1}{2}a\Delta t^{2},\]

the distance traveled at the \(n^\text{th}\) and \((n-1)^\text{th}\) second is

\[\begin{align} S_n &=v_0n + \frac{1}{2}an^{2}\\ S_{n-1} &=v_0(n-1)+\frac{1}{2}a(n-1)^{2}\\ \Rightarrow S_n - S_{n-1} &=v_0n + \frac{1}{2}an^{2}-v_0(n-1)-\frac{1}{2}a(n-1)^{2}\\ &=v_0n + \frac{1}{2}an^{2}-v_0n-v_0-\frac{1}{2}an^2-an-\frac{1}{2}a\\ &=v_0+a\left(n-\frac{1}{2}\right). \end{align}\]

The third equation of motion gives the final velocity of an object under uniform acceleration given the distance traveled and an initial velocity:

\[v^2=v_0^2+2ad.\]

The graph of the motion of the object

Suppose an object is observed from \(t_1\) to \(t_2\). Assume that the object was moving with velocity \(v_0\) at \(t_1\), and it was moving with a velocity of \(v\) at \(t_2\). Below is the graphical proof of the third equation of motion:

The area under the velocity-time graph gives us the displacement. This time, instead of splitting the area into two parts, compute the area of the trapezium directly to find

\[\begin{align} s=\text{Area(trapezium)}&=\dfrac{h(a+b)}2\\ &=\dfrac{(t_2-t_1)(v+v_0)}2\\ &=\Delta t\times\dfrac{(v+v_0)}2\\ &=\dfrac{(v-v_0)}a\times \dfrac{(v+u)}2\qquad \left(\text{since } \Delta t=\dfrac{v-v_0}a\right)\\ \Rightarrow d &=\dfrac{v^2-v_0^2}{2a}. \end{align}\]

Rewriting the above expression, one finds the third equation of motion:

\[v^2=v_0^2+2ad.\]

• Identifying Terms - Position, Velocity, Acceleration

• Average Velocity

• Instantaneous Velocity

• Position-Time Graph - Constant Velocity

• Newton's Laws of Motion

[1] Image from http://www.japantimes.co.jp/sports/2016/01/26/more-sports/track-field/bolt-considering-running-2020-tokyo-games/#.VvEjTMvhVDt.

[2] Image from https://upload.wikimedia.org/wikipedia/commons/4/40/Distancedisplacement.svg under the license Creative Commons Attribution.