Equations of Motion: Derivations, Examples, and Applications

Equations of Motion: Physics’ section on equations of motion helps us in predicting the physical quantities that may change when an object undergoes motion. The various physical quantities associated with motion are the time taken, distance covered, initial velocity, final velocity, and acceleration. Equations of Motion are the relations that relate these physical quantities with each other concerning a uniformly accelerated motion of an object along a straight line.

Motion is basically the change in the object’s position with respect to a fixed reference point. A motion may be uniform or non-uniform. This article will provide you with all the detailed information about Equations of Motion. Continue reading to learn more about Equations of Motion For Uniform Acceleration.

Equations of Motion: Overview

Equations of Motion are the relations that relate the physical quantities like distance travelled, time taken, initial velocity, final velocity and acceleration with each other to facilitate the calculations with regards to a uniformly accelerated motion of an object along a straight line.

For example, let us take the case of aly falling body dropped from a height. It has zero initial velocity but has acceleration which is nothing but the acceleration due to gravity that is uniform. Due to this reason, the velocity of the body will keep on increasing and will achieve a different value while reaching the ground.

It will take some time and would have covered some distance while falling from the height. Here, it is to be considered that the motion of the body is along a straight line. So, with the help of the equations of motion, we can calculate the value of the unknown physical quantities if sufficient data is provided.

Definition of Equations of Motion

Equations of Motion can be defined with respect to the components it relates.

When an object moves along a straight line with uniform acceleration, we can relate its velocity, acceleration and distance travelled in a particular time with a set of equations. These equations are called equations of motion. There are three equations of motion that are as listed below:

1.\(v = u + at\)
2.\(s = ut + \frac{1}{2}a{t^2}\)
3.\({v^2} – {u^2} = 2as\)
We will derive each of them by both graphical and algebraic method.

Derivation of First Equation of Motion by Graphical Method

The First Equation of Motion that relates the velocity and time of the object in uniformly accelerated motion along a straight line is derived graphically as given below:

The velocity-time graph of an object that is moving with uniform acceleration along a straight line is given below:

Here, the position of the object at time \(t = 0\) is represented by the letter \(A.\) In time \(t,\) the object reaches position \(B.\)
So, from the graph, we can conclude the following:
The time taken by the object is represented by \(OD = t,\) which is also equal to \(AC\) and \(EB.\)
The initial velocity of the object is represented by \(OA = u,\) which is also equal to \(DC.\)
The final velocity of the object is represented by \(OE = v,\) which is also equal to \(DB.\) We know that the slope of the graph gives the acceleration of the object.
Thus, the acceleration is \(a = \,{\rm{slope}}\, = \,\frac{{CB}}{{AC}}\)
From the graph, we get \(CB = DB – DC = v – u\) and \(AC = t.\) Putting these values in the above equation, we get,
\(a = \frac{{v – u}}{t}\)
\(\therefore at = v – u\)
\(\therefore u + at = v\)
Or
\(v = u + at\)
which is the required relation connecting the velocity and the time of travel of the object in motion. Thus, the relation \(v = u + at\) is also known as velocity \( – \) time relation.

Derivation of First Equation of Motion by Algebraic Method

The First Equation of Motion can be derived algebraically also. We know that acceleration is the rate of change of velocity.

\(\therefore a = \frac{{v – u}}{t}\)
\(\therefore at = v – u\)
\(\therefore u + at = v\)
Or
\(v = u + at\)
which is the required first equation of motion that is also known as velocity \( – \) time relation.

Derivation of Second Equation of Motion by Graphical Method

The Second Equation of Motion that relates the position and time of travel of the object in uniformly accelerated motion along a straight line is derived graphically as given below:

The velocity \( – \) time graph of an object that is moving with uniform acceleration along a straight line is as given below:

Here, the position of the object at time \(t = 0\) is represented by the letter \(A.\) In time \(t,\) the object reaches position \(B.\)
So, from the graph, we can conclude the following:
The time taken by the object is represented by \(OD = t,\) which is also equal to \(AC\) and \(EB.\)
The initial velocity of the object is represented by \(OA = u,\) which is also equal to \(DC.\)
The final velocity of the object is represented by \(OE = v,\) which is also equal to \(DB.\)
We know that the distance travelled by the object is given by the area of the space enclosed between the velocity \( – \) time graph and the time axis.
Thus, the distance travelled is \(s = {\rm{area\;of\;the\;figure}}\;OABD\)
\(\therefore s = {\rm{area\;of\;rectangle}}\;OACD + {\rm{area\;of\;triangle}}\;ABC\)
\(\therefore s = \left( {OA \times AC} \right) + \left( {\frac{1}{2} \times CB \times AC} \right)\)
Here, \(OA = u,\,AC = t,\,CB = DB – DC = v – u\) and \(AC = t.\) Putting these values in the above equation, we get,
\(s = ut + \frac{1}{2} \times \left({v – u} \right) \times t\)
But from the velocity – time relation, we get, \(v – u = at.\) Putting this value in the above equation, we get,
\(s = ut + \frac{1}{2} \times at \times t\)
\(\therefore s = ut + \frac{1}{2}a{t^2}\)
which is the required relation connecting the position and the time of the object in motion. Thus, the relation \(s = ut + \frac{1}{2}a{t^2}\) is also known as position \( – \) time relation.

Get Examples of Equation of Motion for Acceleration

Derivation of Second Equation of Motion by Algebraic Method

The Second Equation of Motion can be derived algebraically also.
We know that velocity is the rate of change of displacement.
Thus, the displacement is the product of velocity and time taken.
\(\therefore \,\,{\rm{displacement}} = {\rm{velocity}} \times {\rm{time}}\)
\(\therefore s = v \times t\)
This holds good when the velocity is constant.
But if the velocity is not constant, then the average velocity is considered (for uniformly accelerating motion) in place of velocity. Hence, the displacement in the case of uniformly accelerating motion becomes the product of average velocity and time.
\(\therefore {\mkern 1mu} \,{\mkern 1mu} {\rm{displacement}} = {\rm{average}}\,{\mkern 1mu} {\rm{velocity}} \times {\rm{time}}\)
\(\therefore s = {v_{av}} \times t\)
Average velocity is given by the relation \({v_{av}} = \frac{{u + v}}{2}.\) Putting this value of average velocity in the above relation, we get,
\(s = \frac{{u + v}}{2} \times t\)
But from the first equation of motion, we get \(v = u + at.\) Putting this value of v in the above relation, we get,
\(s = \frac{{u + \left({u + at} \right)}}{2} \times t\)
\(\therefore s = \frac{{2u + at}}{2} \times t\)
\(\therefore s = \left({\frac{{2u}}{2} + \frac{{at}}{2}} \right) \times t\)
\(\therefore s = \left({u + \frac{{at}}{2}} \right) \times t\)
\(\therefore s = ut + \frac{1}{2}a{t^2}\)
which is the required second equation of motion that is also known as position \( – \) time relation.

Derivation of Third Equation of Motion by Graphical Method

The Third Equation of Motion that relates the position and velocity of the object in uniformly accelerated motion along a straight line is derived graphically as under:

The velocity \( – \) time graph of an object that is moving with uniform acceleration along a straight line is as under:

Here, the position of the object at time \(t = 0\) is represented by the letter \(A.\) In time \(t,\) the object reaches position \(B.\)
So, from the graph, we can conclude the following:
The time taken by the object is represented by \(OD = t,\) which is also equal to \(AC\) and \(EB.\)
The initial velocity of the object is represented by \(OA = u,\) which is also equal to \(DC.\)
The final velocity of the object is represented by \(OE = v,\) which is also equal to \(DB.\)
We know that the distance travelled by the object is given by the area of the space enclosed between the velocity – time graph and the time axis.
Thus, the distance travelled is \(s = {\rm{area\;of\;the\;trapezium}}\;OABD\)
\(\therefore s = \frac{{{\rm{sum\;of\;parallel\;sides}}}}{2} \times {\rm{height}}\)
\(\therefore s = \frac{{OA + DB}}{2} \times OD\)
Here, \(OA = u,\,DB = v\) and \(OD = t.\) Putting these values in the above equation, we get,
\(s = \frac{{u + v}}{2} \times t\)
But from the velocity – time relation, we get, \(t = \frac{{v – u}}{a},\) Putting this value in the above equation, we get,
\(s = \frac{{v + u}}{2} \times \frac{{v – u}}{a}\)
\(\therefore s = \frac{{{v^2} – {u^2}}}{{2a}}\)
\(\therefore 2as = {v^2} – {u^2}\)
or \({v^2} – {u^2} = 2as\)
which is the required relation connecting the position and the velocity of the object in motion. Thus, the relation \({v^2} – {u^2} = 2as\) is also known as position – velocity relation.

Derivation of Third Equation of Motion by Algebraic Method

The Third Equation of Motion can be derived algebraically also.
We know that velocity is the rate of change of displacement.
Thus, the displacement is the product of velocity and time taken.
\(\therefore {\rm{displacement}} = {\rm{velocity}} \times {\rm{time}}\)
\(\therefore s = v \times t\)
This holds good when the velocity is constant.
But if the velocity is not constant, then the average velocity is considered (for uniformly accelerated motion) in place of velocity. Hence, the displacement in the case of uniformly accelerated motion becomes the product of average velocity and time.
\(\therefore {\rm{displacement}} = {\rm{average}}{\mkern 1mu} \,{\rm{velocity}} \times {\rm{time}}\)
\(\therefore s = {v_{av}} \times t\)
Average velocity is given by the relation \({v_{av}} = \frac{{u + v}}{2}.\) Putting this value of average velocity in the above relation, we get,
\(s = \frac{{u + v}}{2} \times t\)
But from the first equation of motion, we get \(t = \frac{{v – u}}{a}.\) Putting this value in the above equation, we get,
\(s = \frac{{v + u}}{2} \times \frac{{v – u}}{a}\)
\(\therefore s = \frac{{{v^2} + {u^2}}}{{2a}}\)
\(\therefore 2as = {v^2} -{u^2}\)
or
\({v^2} – {u^2} = 2as\)
which is the required third equation of motion that is also known as position \( – \) velocity relation.

Names of the Equations of Motion

Equations of Motion have a specific name based on all physical quantities they are relating by the equation. The name of the three equations of motion are as under:

1 The First Equation of Motion is, \(v = u + at\) is known as velocity \( – \) time relation.
2 The Second Equation of Motion is, \(s = ut + \frac{1}{2}a{t^2}\) is known as position \( – \) time relation.
3 The Third Equation of Motion is, \({v^2} – {u^2} = 2as\) is known as position \( – \) velocity relation.

Solved Examples on Equations of Motion

Q.1. A body is accelerating at a constant rate of \(10{\mkern 1mu} \,{\rm{m}}\,{{\rm{s}}^{ – 2}}.\) If the body starts from rest, how much distance will it cover in \(5{\mkern 1mu} \,{\rm{s}}\)?
Ans: Given that,
The acceleration of the body is \(a = 10{\mkern 1mu} \,{\rm{m}}\,{{\rm{s}}^{ – 2}}\)
The initial velocity of the body is \(u = 0{\mkern 1mu} \,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
The time taken by the body is \(t = 5{\mkern 1mu} \,{\rm{s}}\)
The distance covered by the body is \(s = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2} \times 10 \times {5^2} = 125{\mkern 1mu} \,{\rm{m}}\)
Thus, the body covers a distance of \(125\,{\mkern 1mu} {\rm{m}}{\rm{.}}\)

Q.2. A car accelerates uniformly from \(36\,{\mkern 1mu} {\rm{km}}\,{\mkern 1mu} {{\rm{h}}^{ – 1}}\) to \(54{\mkern 1mu} \,{\rm{km}}{\mkern 1mu} \,{{\rm{h}}^{ – 1}}\) in \(5{\mkern 1mu} \,{\rm{s}}.\) Calculate the acceleration of the car as well as the distance covered by it in that time.
Ans: Given that, The initial velocity of the car is \(u = 36{\mkern 1mu} \,{\rm{km}}\,{\mkern 1mu} {{\rm{h}}^{ – 1}} = 10{\mkern 1mu} \,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
The final velocity of the car is \(v = 54\,{\rm{km}}{\mkern 1mu} \,{{\rm{h}}^{ – 1}} = 15{\mkern 1mu} \,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
The time taken by the car is \(t = 5{\mkern 1mu} \,{\rm{s}}\)
The acceleration of the car is \(a = \frac{{v – u}}{t} = \frac{{15 – 10}}{5} = 1\,{\rm{m}}\,{{\rm{s}}^{ – 2}}\)
The distance covered by the car is \(s = ut + \frac{1}{2}a{t^2}\)
\(\therefore s = 10 \times 5 + \frac{1}{2} \times 1 \times {5^2} = 50 + 12.5 = 62.5{\mkern 1mu} \,{\rm{m}}\)
Thus, the acceleration of the car is \(1\,{\mkern 1mu} {\rm{m}}\,{{\rm{s}}^{ – 2}}\) and it covers \(6.25{\mkern 1mu} \,{\rm{m}}\) in the given duration of time.

Applications of Equations of Motion

Equations of Motion find many applications in the real world. A few of them are listed below:

1. We can predict the distance travelled by the body if its initial and final velocities are known along with its acceleration without even knowing the time for which the body travelled.
2. We can predict the time of motion of the body just by knowing its initial and final velocities along with its acceleration, even without knowing how far the body has travelled.
3. We may also know the distance travelled by the body just by knowing its initial velocity, acceleration and time for which it travelled without knowing what would have been its final velocity while the body stopped.
4. Equations of motion also help us to calculate the initial velocity of the body just by knowing its final velocity, acceleration and time for which the body travelled.
5. Equations of motion help us to get the value of the final velocity the body will achieve after the given time if its initial velocity and acceleration are known to us.

Thus, the equations of motion help us to calculate the quantities related to them just by knowing any three of the quantities in them and therefore help in getting the rest of the physical quantities.

Summary

Equations of Motion are the relations that relate the physical quantities such as time taken, distance travelled, initial velocity, final velocity and acceleration with each other to facilitate the calculations about the uniformly accelerated motion of an object along a straight line. It has a particular name based on physical quantities they are related by the equation. The three equations of motion are velocity-time relation, position-time relation, and position velocity relation. It is also important to note that to apply equations of motion the object should be uniformly accelerated and should be travelling along a straight line.

FAQs on Equations of Motion

Q.1: What is the other name used for the equations of motion?
Ans: The first equation of motion \(v = u + at\) is also known as velocity \( – \) time relation, the second equation of motion \(s = ut + \frac{1}{2}a{t^2}\) is also known as position \( – \) time relation and the third equation of motion \({v^2} = {u^2} = 2as\) is also known as position \( – \) velocity relation.

Q.2: What are the conditions required to apply equations of motion?
Ans: The conditions to apply equations of motion are that the body should be uniformly accelerated and should be travelling along a straight line.

Q.3: What are the three equations of motion?
Ans: The three equations of motion are \(v = u + at,\,s = ut + \frac{1}{2}a{t^2}\) and \({v^2} – {u^2} = 2as.\)

Q.4: Can equations of motion be applied on aly falling body?
Ans: Equations of motion can be applied on aly falling body, as the acceleration of the body is uniform, which is the acceleration due to gravity, and the body travels along a straight line.

Q.5: What will happen to the equations of motion when the acceleration becomes zero?
Ans: When the acceleration becomes zero, the equations of motion will become \(v = u,\,s = ut\) and \({v^2} = {u^2}\) which also means \(v = u.\)

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