Laws of Motion – Explanation, Derivation and FAQs

Laws of Motion – Explanation in Detali

Newton’s three laws of motion are the physical laws that govern the relationship between a body and the forces acting on it, and its state of motion in response to these forces.


We all have learned about the concepts of displacement, velocity, acceleration, and derived relation between these quantities. In this article, we are going to understand the cause of motion. A body may or may not start moving when the external force is applied to the body. We shall also learn the effects produced by the forces on the body.


Derivation of First Equation of Motion

The first equation of motion can be easily understood by the graph method, for that, let’s look at the graph drawn below:

Let’s suppose that you are walking on the roof of your house at 5 m/s and suddenly the light goes off in your house, you start running out of your house because of fear at a speed of 20 m/s. Now, what is the difference in your velocity? Its is given by:

v = 20 m/s, u = 5 m/s, i.e.,  v - u = 20 - 5 = 15 m/s. 

So, from the graph: OC - OD = CD = Difference in velocity = 15 m/s.

Now, if you are moving with a uniform acceleration ‘a’, we also know that change in velocity per unit time is acceleration. 

From here, a  = slope of the graph, which is given as:

a  = AB/AD

And, AB = BE - AE

We can see that BE = v and AE = u

Putting the value of AB, we get:

a  = (v - u), and also AD = t

So, we get the formula for acceleration as:

a = change in velocity per unit time =  (v  - u)/t

Or

v- u = at => v = u + at, is the required derivation of the first equation of motion.

If the man was sitting on the roof, which means his u = 0 m/s, then the equation can be re-written as:

v  = at….(1)


Derivation of Second Equation of Motion

The second equation of motion describes the Position of an object with respect to time. Let’s derive the second equation of motion with the help of the graph drawn below:


(Image will be Uploaded soon)


We know that the area enclosed by the graph in Fig.a  is equal to the distance traveled by the body. So, for finding the distance traveled, we need to find the area of trapezium.

Now, finding the area of trapezium OPQS:

s = Area of Rectangle OPRS + Area of triangle QPR

= (OP x PR) + 1/ 2 * QR * PR…(a)

Here, 

OP = u

PR = t

QR = v - u

On putting these values in equation (a), we get:

= (u x t) + 1/2 *  (v - u) * t

We know that  v - u = at

So, 

s = u * t  + 1/2 (at) * t = ut + 1/2 at2, is the required second equation of motion.

Here, s is the distance traveled between two points.


Derivation of Third Equation of Motion

The third law of motion describes the velocity-time relationship. Now, let’s derive the third law of motion with the help of Fig.b:

From Fig.b, the area of trapezium can be calculated as:

Area of trapezium OPQS = ½ * (OP + QS) * OS = ½ * (RS + QS) * OS….(2)

Acceleration (a) = slope of velocity-time graph PQ, which is given by:

a = QR/PR = (QS - RS)/OS

Or

OS = (QS - RS)/a…..(3)

Now, putting the value of (3) in (2), we get:

OS =  ½ * (RS + QS) * (QS - RS)/a

 OS = 1/2a (QS2 - RS2)

Here,

OS = s

RS = u

QS = v

So, we get the third equation of motion as:

s = 1/2a (v2 - u2)

Or

2as = (v2 - u2)

Or

v2 = u2 + 2as


Newton’s Second Law Motion

This law states the relationship between the linear momentum of a body and the force applied to it. This law states that the change of linear momentum of a body per unit time varies directly with the force applied to it and this change takes place in the direction of the applied force. It means that when the force is bigger, the linear momentum of the body changes quickly. It is given by:

F ∝ dp/dt….(2)

And, p = mv

Where,

p = linear momentum of the body

t =  time taken

dp/dt = rate of change of linear momentum

m =  mass of the body

v = velocity of the body

Now, equating the value of p in eq (2), we get:

 F∝mv/dt ⇒F ∝mdv/dt

F∝mvdt⇒F∝mdv dt

Removing the sign of proportionality constant, the formula for second law of motion becomes:

                        F = k mdv/dt

We know that  v = at from eq (1), we get:

                        F = k ma→

, is the required derivation of Newton’s second law of motion.

Where a  =⟶dvdt  represents the acceleration of the body.


Laws of Motion – Explanation, Derivation and FAQs

Isaac Newton discovered the Laws of Motion. Newton’s laws of motion explain the connection between a physical object and the forces acting upon it. By understanding the concept, it provides all the information about the basis of modern physics. After the development of the three laws of motion, Newton revolutionized science. Newton’s three laws together with Kepler’s Laws explained the concept of planets moving in elliptical orbits rather than in circles. Although Newton’s laws of motion may seem so general today, they were considered revolutionary centuries ago. Newton’s three laws of motion help someone understand how objects behave when standing still, when moving and when forces act upon them. Sir Newton’s three laws are Newton’s First Law: Inertia, Newton’s Second Law: Force, Newton’s Third Law: Action & Reaction.


Newton’s First Law

Newton’s First Law states that ‘An object at rest remains at rest, and an object in motion remains in motion at a constant speed and in a straight line unless acted on by an unbalanced force’. This law describes that every object will remain at rest or in uniform motion in a straight line unless forced to change its state by the action of an external force. The tendency of an object to resist changes in a state of motion is inertia. There will be a net force on an object if all the external forces cancel each other out. If this happens, then the object will maintain a constant velocity. If the Velocity remains at zero, then the object remains at rest. If any external force acts on an object, the velocity will change because of the force applied. An example of the First law of motion is : The motion of a round ball falling down through the atmosphere, A rocket being launched up into the atmosphere.


Newton’s Second Law

Newton’s Second Law states that ‘The acceleration of an object depends on the mass of the object and the amount of force applied’. This law defines a force to be equal to a change in momentum (mass times velocity) per change in time. Momentum is described as the mass m of an object times its velocity V. Example of the second law of motion is: An aircraft’s motion resulting from the aerodynamic force, aircraft thrust and weight.


Newton’s Third Law

Newton’s Third Law states that ‘Whenever one object implies a force on a second object, the second object implies an equal and opposite force on the first’. This law defines that for every action in nature there is an equal and opposite reaction. If object A applied a force on object B, object B will also apply an equal and opposite force on object A. In other words, it can be said that forces result from interactions. An example of the third law of motion is: The motion of a jet engine produces thrust and hot exhaust gasses flow out the back of the engine, and a thrusting force is produced in the opposite direction.


Derivation of First Equation of Motion

Let’s Consider a body of mass m having initial velocity u.

Let after time be t its final velocity becomes v due to uniform acceleration a.

Now it is defined as:

Acceleration = Change in velocity / Time taken

Acceleration = (Final velocity - Initial velocity) / Time taken

 a = v - u / t

 a t = v - u

or  v = u + a t

This describes the first equation of motion.


Derivation of Second Equation of Motion

As it is defined, the Second equation of motion: s = u t + 1/2 at2

Let’s take the distance traveled by the body be s.

Now:

Distance = Average velocity x Time

Also, Average velocity = (u + v) / 2

Therefore, Distance (t) = (u + v) / 2 t       .eq.(1)

Again from first equation of motion:

v = u + at

Substituting this value of v in eq.(1), we get

s = (u + u + a t) / 2 t

s = (2 u + a t) / 2 t

s = (2 u t + a t2) / 2

s = (2 u t / 2) + (a t2 / 2)

or  s = u t + (1/2) a t2

This describes the second equation of motion.


Derivation of Third Equation of Motion

As it is defined that the third equation of Motion:  v2 = u2 +2 a s

Now,

v = u + a t

v - u = a t

or  t = (v - u) / a                   eq.(2)

Also ,

Distance = average velocity x Time

Therefore, s = [(v + u) / 2] x [(v - u) / a]

s = (v2 - u2) / 2 a

2 a s = v2 - u2

or  v2 = u2 + 2 a s

This describes the third equation of motion.

FAQs (Frequently Asked Questions)

1. Define Newton’s First Law of Motion?

Newton’s first law of motion states that ‘An object at rest remains at rest, and an object in motion remains in motion at a constant speed and in a straight line unless acted on by an unbalanced force’. First law of motion describes that things cannot start, stop, or change direction all by themselves, and it requires some force from the outside to cause such a change. This property of an object or a body to resist changes in its state of motion is called inertia. And thus the first law of motion is also known as the law of inertia.

2. Define Newton's Second Law of Motion?

Newton’s second law of motion states that ‘The acceleration of an object depends on the mass of the object and the amount of force applied’. Second law of motion defines what will happen to the massive body when acted upon by an external force. This law of motion states that the force acting on the body is equal to the product of its mass and acceleration. This law also describes precisely how much an object will accelerate for a given net force.

3. Define Newton’s Third Law of Motion?

Newton’s Third Law of motion states that ‘Whenever one object implies a force on a second object, the second object implies an equal and opposite force on the first’. Third law describes that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. The law is also known as the law of action and reaction. Third law is essential in analyzing problems of static equilibrium, where all forces are balanced, but it also applies to bodies in uniform or accelerated motion.

4. Describe the real life example for three laws of motion?

Example of First law of motion is: The motion of a round ball falling down through the atmosphere. 

Example of the second law of motion is: An aircraft’s motion resulting from the aerodynamic force, aircraft thrust and weight. 

Example of the third law of motion is : The motion of a jet engine produces thrust and hot exhaust gasses flow out the back of the engine, and a thrusting force is produced in the opposite direction.

5. From where can a student find the study materials for the concept Laws of Motion?

Students can find everything they need on the vedantu app . The study materials are created by professionals in the field and the content is accurate and reliable. These study materials are completely free and there are no charges at all. All students just have to sign in and then they will be able to download what they want in pdf format. Students can find various study materials related to Laws of motion. Every student can take advantage of these free resources that will surely help you ace your exams. 

6. What are the Real-Life Applications of Newton’s Three Laws of Motion?

Let’s discuss the real-life application of three laws of motion:

  • First Law of Motion: A ball falling from the atmosphere.

  • Second Law of Motion: If you push the car and the truck with the same force, the car will have greater acceleration.

  • Third Law of Motion: While doing Bungee Jumping, the earth exerts a gravitational force on the mass of the jumper, while at the same time the bungee reacts by exerting an opposite but equal force.

7. What Three Physical Quantities are Used in Equations of Motion?

The following quantities are used in the equation of motion:

  1. Displacement

  2. Velocity: Initial and final 

  3. Acceleration

  4. Time

8. A Ball is Thrown Vertically Upwards with a Speed of 19.8 m/s From the Roof of the Apartment Building and Returns to the Ground in 6 Seconds. Find the Height of the Building.

Given t = 6 s, a = 9.8ms-2, u = - 19.8 m/s

Here, u = - ve because we are taking downward vertical motion of the ball from the point of projection up to ground.

We know that s = ut + 1/2at2, now putting the above values in this equation, we get:

s = - 19.8 x 6 + ½ * 9.8 * 62

Height of the roof = 57.6 m

9. Who Discovered Gravity?

The gravity was discovered by Sir Isaac Newton.

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