## 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:

(image will be uploaded soon)

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 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 at^{2}, 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:

= ½ * (RS + QS) * (QS - RS)/a

OS = 1/2a (QS^{2} - RS^{2})

Here,

OS = s

RS = u

QS = v

So, we get the third equation of motion as:

s = 1/2a (v^{2} - u^{2})

Or

2as = (v^{2} - u^{2})

Or

v^{2} = u^{2} + 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

\[\overrightarrow{v}\] = velocity of the body

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

\[F \propto \frac{mv}{dt} \Rightarrow F \propto \frac{mdv}{dt}\]

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

F = \[k \frac{mdv}{dt}\]

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

F = k \[\overrightarrow{ma}\], is the required derivation of Newton’s second law of motion.

Where a = \[\overrightarrow{dv}{dt}\] represents the acceleration of the body.

Question 1: What are the Real-Life Applications of Newton’s Three Laws of Motion?

Answer: 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.

Question 2: What Three Physical Quantities are Used in Equations of Motion?

Answer: The following quantities are used in the equation of motion:

Displacement

Velocity: Initial and final

Acceleration

Time

Question 3: 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.

Answer: 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 + 1/2at^{2}, now putting the above values in this equation, we get:

s = - 19.8 x 6 + ½ * 9.8 * 6^{2}

Height of the roof = 57.6 m

Question 4: Who Discovered Gravity?

Answer: The gravity was discovered by Sir Isaac Newton.