Category: Physics Notes PPT

Example: What is the intensity of light incident normal to a circular surface of radius 5 cm from a 100 W source of light?

Solution:

r = 5 cm = 5 × 10–2 m, P = 100 W, [I]=?

[I = frac{P}{{pi {r^2}}} = frac{{100}}{{pi times {{(5 times {{10}^{–2}})}^2}}} = 1.27 times {10^4}W{m^{–2}}]

Question: The intensity of a light source with power P on a circular surface of radius r placed at a certain distance is [I]. If at the same distance from another source, if the intensity on a circular surface of double the radius, is 4[I], then what is the power of the second source?

Options:

(a) 4 P

(b) 16 P

(c) 2 P

(d) P / 4

Answer: (b)

For an object exhibiting a circular motion, there are always some parameters to describe its nature. 

 

If we talk about a particle’s velocity, which is an angular velocity, that remains constant throughout the motion; however, angular acceleration makes two types of components and they are tangential and radial acceleration.

 

Tangential acceleration acts tangentially to the direction of motion of a particle and remains perpendicular to the direction of the radial component. Now, we’ll discuss the tangential and radiation acceleration formula.

 

Tangential Acceleration and Centripetal Acceleration Formula

Tangential acceleration meaning is a measure of how the tangential velocity of a point at a given radius varies with time. Tangential acceleration is just like linear acceleration; however, it’s more inclined to the tangential direction, which is related to circular motion.

 

If we talk about the narrow gap between the centripetal acceleration, which is an acceleration that acts towards the center of the circle along which the body or a particle is creating a circular motion.

 

We can see there is a narrow line of difference between the two types of acceleration, and that the difference lies in the way the acceleration acts on the particle in a circular motion.

 

Now, let’s discuss the tangential acceleration equation followed by the centripetal acceleration.

 

Tangential Acceleration Formula

Let’s suppose that you and your friends are playing with a string. You are in the middle of the string and your friends have joined the string from hand to hand and are moving with high speed or changing speed in a circular motion.

 

Here, we are talking about angular velocity and we know that change in the velocity is called acceleration, which is angular acceleration. So, we can write the first derivative of angular velocity concerning time for angular acceleration. 

 

Acceleration Tangential Formula 

Here, we aim to describe the tangential acceleration formula, so we will focus more on it, as our article relies on the same. 

 

Now, writing the tangential acceleration equation in the following manner:

    

Tangential acceleration [a_{t} = r times frac{text{d}omega}{text{d}t}]……(1)

 

So, we denote the tangential acceleration with a subscript ‘ct’ along with the English letter ‘a’. 

  Here, [frac{text{d}omega}{text{d}t}] = angular acceleration

r = is the radius of the circle

 

Since the motion talks about the position of a particular object that’s why we call ‘r’ as the radius vector.

 

We also know that the angular velocity can be written, so we can rewrite the above equation (1) to get the Tangential Acceleration Formula Circular Motion in a new form:

at = r𝛼….(2)

 

Centripetal Acceleration Formula

The centripetal acceleration of an object making a circular motion with a circle ‘r’ and having a speed ‘v’ in meter per second is given by the following centripetal acceleration equation:    

 

                      aC = v2/r

 

So, we denote the centripetal acceleration with a subscript ‘c’ along with the English letter ‘a’. 

 

Now, we will discuss the radial and tangential acceleration formula in detail.

 

Tangential and Radial Acceleration Formula 

We already discussed the acceleration tangential formula in the above context, while talking about the narrow difference between the centripetal and tangential acceleration, we also saw a minor difference between tangential acceleration and the centripetal acceleration formula.

              

Now, let’s discuss the radial acceleration:

 

Radial Acceleration

We define radial acceleration as the component that points along the radius vector, the position vector that points from a center, usually of rotation, and the position of the particle that is accelerating.

 

The formula for radial acceleration is given by: 

                   ar = v2/r …..(3)

 

Here, we can see the term ‘r’ or the radius vector has a difference in the tangential acceleration and the centripetal acceleration formula. Also, we notice that the centripetal acceleration and the radial acceleration have the same formula.

 

 

Finding Tangential Acceleration

Now, we will look at one problem to find the tangential acceleration of an object. 

 

Example: 

If an object is experiencing a circular motion, then what will be its tangential acceleration? Also, determine the overall acceleration of the object.

 

Answer: 

The overall acceleration of an object is given by the following equation:

 

[vec{a_{(total)}} = vec{a_{(r)}} + vec{a_{(t)}}]

Now, tangential acceleration can be determined by subtracting the radial component acceleration from the overall acceleration in the following manner:   

 

 [vec{a_{(t)}} = vec{a_{(total)}} – vec{a_{(r)}}] 

 [vec{a_{(t)}} theta cap = vec{a_{(total)}} – vec{a_{(r)}} r(cap)]

If we wish to find out the total acceleration in the modulus function, we have the following equation:

 

[vec{a_{(total)}} = |vec{a_{(total)}}| = sqrt{a_r^2}+a_t^2] 

So, the total acceleration is the square root of the sum of the squares of the radial and tangential acceleration.

 

What are the possibilities for the Value of Tangential Acceleration?

The result of tangential acceleration may have the following three possibilities:

1. The value of tangential acceleration can be greater than zero, that is, positive. This generally happens when the magnitude of the velocity vector increases with time, that is, the body has accelerated motion.

2. The value of tangential acceleration can be less than zero, that is, negative. This happens when the magnitude of the velocity vector decreases with time, that is when the object has slowed or decelerated motion.

3. The value of tangential acceleration can be equal to zero. This happens when the magnitude of the velocity vector remains constant, that is, the object is in uniform motion.

 

Summary of Tangential Acceleration

The object has a constant speed when it is executing circular motion around a circle of fixed radius in case of uniform circular motion. But if in case, the speed of the object is not constant, this is, changing with time then there will be an additional acceleration applied on the o
bject, that is known as tangential acceleration, and is applied in the direction of the tangent of the circle. The tangential acceleration is calculated as the product of the radius of the circle and angular acceleration. The rate of change of angular velocity is denoted by angular acceleration. The centripetal acceleration is the radial component of the acceleration and is calculated as the square of velocity divided by the radius of the circle. Centripetal acceleration is directed towards the center of the circular path, that is, radially inwards. Both these accelerations, that is, centripetal acceleration and tangential acceleration are mutually perpendicular to each other. Therefore, the total acceleration acting on a particular object is the sum of both the tangential acceleration and the centripetal acceleration.

 

For example, a particle moving in a circular path of radius 2m has a linear speed equal to the square of time in seconds. The linear speed is given in meters per second. Calculate the radial acceleration and the tangential acceleration on the particle at t = 2s. Also, compute the net acceleration acting on a particle.

Solution: Since, the linear speed of the particle is the square of time in seconds, therefore, the linear speed of the particle at t = 2s will be the square of 2, that is, 4 meters per second. 

The radial acceleration of a particle is calculated as the square of velocity divided by the radius of the circle. Since the velocity as calculated above is 4 m/s, therefore the square of 4 will be 16m/s and the radius of the circle given in the question is 2m. Thus, radial acceleration will be 16 divided by 2, that is, 8 meters per second square.

 

The tangential acceleration of a particle is calculated by differentiation of velocity divided by time. Since the velocity is square of time, therefore on differentiation, we will get 2t. Putting the value of t as 2sec, the acceleration will be the product of 2 and 2, which is 4 meters per second square.

 

Now, for calculating the net acceleration, under the center root of the sum of squares of both the acceleration will be taken. On calculating, this will come out to be 8.944 meters per second square.

Every object with both a magnitude and a direction is referred to as a vector.

A vector can be drawn geometrically as a guided line section with an arrow representing the direction and a length equal to the magnitude of the vector. From the tail to the head, the vector’s orientation is shown. We’ll go over the definition of a vector and some vector formulas with examples in this subject. Let’s take a look at the idea!

Vector Formula 

The Concept of Vector Formula 

In mathematics, a vector is a representation of an object that includes both magnitude and direction.

If two vectors have the same direction and magnitude, they are the same. This means that if we take a vector and transfer it to a different place, we get a new vector. The vector we get at the end of this phase looks like this, and it’s the same vector we had at the start.

In physics, vectors that represent force and velocity are two common examples of vectors. Power and velocity are both acting in the same way. The magnitude of the vector would mean the force’s intensity or the velocity’s related speed. Since displacement is directly attached to distance, distance and displacement are not the same.

An arrow mark is commonly used to represent a vector.

Also, whose length is proportional to the magnitude and whose direction is the same as the quantity. Scaled vector diagrams with values are often used to describe vector quantities. A displacement vector will be described in the vector diagram.

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Some Important Definitions and Vector All Formula

Vector Formula Mathematics

Magnitude

The magnitude of a vector is the length of the vector it is used in the vector formula. The magnitude of the vector a is denoted by |a| For a two-dimensional vector a = (a[_{1}], a[_{2}]), the formula for its magnitude is 

|a| = [sqrt{a_{1}^{2} + a_{2}^{2}}]

And for three-dimensional vector a = (a[_{1}], a[_{2}], a[_{3}]), the formula for its magnitude is 

|a| = [sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}]

Direction

A vector’s direction is often expressed as a counterclockwise angle of rotation around its “tail” from due East.

A vector with a direction of 30 degrees is a vector that has been rotated 30 degrees, counterclockwise relative, to due east using this convention.

Vector Formula Physics

Force 

The vector sum of two or more forces is represented by a resultant force, which is a single force.

Like two forces of magnitudes F1 and F2 function on a particle, the effect is as follows:

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Velocity

The rate of change of an object’s direction is represented by a velocity vector.

The magnitude of a velocity vector indicates an object’s speed, while the vector direction indicates the object’s direction.

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Triangular Law of Additions 

The triangle law of vector addition states that when two vectors are represented as two sides of a triangle of the same order of magnitude and direction, the magnitude and direction of the resulting vector is represented by the third side of the triangle.

As two forces, Vector A and Vector B, function in the same direction, the resulting R is the sum of the two vectors.

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The formula for Triangular law of addition: [bar{R}] = [bar{A}] + [bar{B}] 

 Parallelogram Law of Addition

When two powers, A Vector B Vector formula, are expressed by the parallelogram’s opposite sides, the resultant is represented by the diagonal of a parallelogram taken from the same position.

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The formula for Triangular law of addition: [bar{R}] = [bar{A}] + [bar{B}] 

Vector Subtraction

If two powers, Vector A and Vector B, are acting in the opposite direction, The variance between the two vectors is then used to describe the resultant R.

As a result, the Vector Subtraction formula is  [bar{R}] = [bar{A}] – [bar{B}] 

Note: Any of the concepts and formulae discussed in this vector formula sheet can come in useful when learning about three-dimensional geometry.

A 3D Geometry vector formula sheet is also available on every website.

Examples of Vector Formula

Q.1) Find the Addition and Subtraction of Given Vectors.

1. (2,3,4) and (5,7,8)

2. (6,3,2) and (7,5,3)

Answer:

By using the triangular law of addition the given vectors are,

a) (2,3,4) and (5,7,8)

⇒ {2+5,3+7,4+8}

⇒ {7,10,12}

b) (6,3,2) and (7,5,3)

⇒ {6+7,3+5,2+3}

⇒ {13,8,5}

By using the vector subtraction law the given vector is,

a) (2,3,4) and (5,7,8)

⇒ {2-5,3-7,4-8}

⇒ {-3,-4,-4}

b) (6,3,2) and (7,5,3)

⇒ {6-7,3-5,2-3}

⇒ {-1,-2,-1}

A parent nucleus splits into two or more daughter nuclei to reach the stage of stability. So the unstable nucleus is considered radioactive and while splitting, it loses energy in the form of radiation.

So the whole process initiating from subdivision to loss of energy is the radioactive decay.

The three most common types of decays are:

There are certain radioactive equations for three of these, say, for the gamma decay process, we have the gamma decay formula, proceeding with this, we have the activity of radioactive substance formula. 

Also, the formula for half-life decay helps us determine the time needed for half of the original population of radioactive atoms to decay, which we will understand with the help of the radioactive half-life formula.

Radioactive Decay Equation

As per the activity of radioactive substance formula, the average number of radioactive decays per unit time or the change in the number of radioactive nuclei present is given as:

                                A   =  –  dN/dt

Here,

Also, we understand the following key points from the above radioactivity equation:

 

• The total activity relies entirely on the number of nuclei present, as we can see A dN….(a)

• During the radioactive decay, A decreases with time, as we can see that A ∝ 1/dt…..(b)

Below is the graph representing (a) and (b) definitions:

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Now, let us have a look at alpha, beta, and gamma radioactive equations.

Radioactivity Formula

In the years 1899 and 1900, a British Ernest Rutherford (working at McGill University in Montreal, Canada) and the French Physicist named Paul Villard (working in Paris) did experimental investigations on electromagnetic radiation and separated them into three kinds.

Further, Rutherford named them alpha, beta, and gamma rays depending on the penetration of matter and deflection by a magnetic field.

Here, we will talk about the following three radioactive equations:

• Alpha decay formula

• Beta decay formula

• Gamma decay formula

Alpha Decay Formula

Alpha decay results in the emission of α-particles from the radioactive nucleus.

For example, the alpha decay of  92U238 into   90Th234  is as follows:

 

In the above radioactive decay formula, we notice the following things:

1. The Uranium nucleus emits an α-particle, and therefore, its mass and charge reduced, shown in the following equation:

                 Mass number:  238 –  4 = 234 

                 Charge number: 92 – 2 = 90        

Following this, a new element formed is Thorium (Th).

In general, the radioactivity equation (1) can be represented as:

zXA   ⇾   z – ₂YA-2  +  ₂He⁴ + Q  

Also, we see that after a spontaneous α-decay process, the total mass of  90Th234 and  2He 4 was less than 92U238.          

This means, the total mass-energy (Q) also decreases, equivalent to the difference between the Initial mass-energy and the final mass-energy, stated as:

                             Q = (mx – m – mHe)     

                              Q = (mx – my – mHe). c2   

In this equation, Q is the disintegration energy, which is shared by the daughter nucleus ‘y’ and an alpha particle He.

So, the alpha particle emission occurs in the following manner:

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Beta Decay Formula

Beta decay of Thorium, i.e.,  90Th234 emits a β-particle, where the mass number of the

daughter nucleus remains invariant, while the charge increments by 1, therefore, a new element Palladium 91Pa234  forms in the following way:

90Th234 ⇾  91Pa234 + -1e0 (β-particle)

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Here,

2. The mass number of Palladium is:

               234 – 0 = 234

Its charge number becomes 91 (90+1).

The general radioactivity equation for the beta decay process is:

Z XA ⇾  z+1YA + -1e0 + Q 

 Q = the energy released in β-decay.

Gamma Decay Formula

We know that gamma rays are emitted during the decay of radioactive atomic nuclei and definite subatomic particles. 

These powerful rays are produced by the hottest and most energetic objects in the universe and are present in the electromagnetic spectrum.

 Below is the gamma decay equation of Technetium-99m to Technetium-99:

 43Tc99m      →      43Tc99      +   0γ0   (Gamma radiation) ….(2)

Another e
xample that initiates from the β-decay of 27Co60 turning into an exciting 28Ni60 nucleus, the radioactive decay equation for the same with the energy released is as follows:

27Co60  ⇾ 28Ni60**   + -1e0……(3)

So, when this exciting nucleus reaches the ground state, as a result, gamma rays are emitted with a release of energy in Mega electron Volts.

  28Ni60** ⇾ 28Ni60* + Eγ ( = 1.17 MeV)……(4)

  28Ni60* ⇾ 28Ni + Eγ (= 1.33 MeV)……(5)

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Radioactive Half Life Formula

The formula for half-life decay is:

[N(t)=N_{0}(frac{1}{2})^{frac{t}{t_{1/2}}}]……(6)

Here,

• N (t) is a function of time, which shows the amount of substance remaining after the decay in a given time.

• N is the initial quantity of the substance

• t  is the time elapsed, and

• t1/2  is the half-life of the decaying component

Definition of the Half-Life:

When half of the radioactive atom undergoes the decay process, the time needed for a quantity to reduce to half of its initial value is the half-life. When talking about the decay of half of the radioactive atoms, the time taken is the radioactive half-life. 

For this, we have a radioactive half-life formula:

[t_{1/2}=frac{0.693}{lambda }]

Here, λ is the decay constant. 

Now, let us understand the decay constant formula:

Let’s suppose that ‘N’ is the size of a population of radioactive atoms at a given time ‘t,’ and dN is the amount by which the population of the radioactive atom decreases in time dt; therefore,  the rate of change is given by the following equation:                          

dN/dt     = – λ N,  (λ  = decay constant)

Radioactive Half-Life Table

Radioactive Half-Life Graph

The graph of the above table is as follows:

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Radioactive Half-Life Formula Derivation

Let’s describe equation (6) in an exponential form:

[N(t)=N_{0}(frac{1}{2})^{frac{t}{t_{1/2}}}]

[N(t)=N_{0}e^{frac{-t}{tau}}]

Here,

τ =  A mean lifetime of the decaying quantity, which is positive 

λ is also positive

The three parameters (t1/2, τ , and λ are all directly related to each other in the following way:

[t_{1/2}=frac{ln(2)}{lambda }]  =   (ln (2). τ

We know that the value of the natural logarithmic of 2 is 0.693, so rewriting equation (7):

[t_{1/2}=frac{0.693}{lambda }]

This is the required radioactivity formula for radioactive half-life.

Conclusion

A phenomenon in which a ‌heavy‌ ‌unstable element‌ ‌disintegrates‌ ‌itself‌ into two or more daughter nuclei ‌without‌ ‌being‌ forced‌ ‌by‌ ‌any‌ ‌external‌ ‌agent‌ ‌to‌ ‌do‌ ‌so.‌ The process involves the emission of the following particles:

• α particle

• β particle

• 𝛾 particle

While discussing the thermal properties of materials, specific heat is a concept that very few people discuss and are aware of it. We use the specific heat concept to understand how much heat has to be supplied to raise the temperature of an object by a degree in kelvin or celsius. The amount of heat required to raise the temperature of the substance is known as the specific heat. In other words, the specific heat is the amount of heat required to change the temperature of the material or substance under consideration.

In this article, we will discuss what is specific heat, what is the formula of specific heat along with a small derivation, and solve numerical problems. 

Specific Heat Equation

We might have noticed in our day-to-day life if we consider heating or boiling water in two vessels of different volumes, the amount of heat required to boil the water at a certain time will completely depend on the volume of the water. In other words, if one vessel is containing more water than the other vessel, then the time taken by the vessel with more volume will be more than the other vessel. This implies that the amount of heat required to raise the temperature of any given substance is different and depends on the volume of the substance, this particular amount of heat required to vary the temperature is known as the specific heat or specific heat capacity.

We know that the term specific heat in thermodynamics refers to the ratio of the quantity of heat that we require to raise the temperature of a substance by one degree celsius that we need to increase the temperature of an equivalent mass of liquid (say water) by one-degree celsius. At the same time, we use the term specific heat for a more conventional reason to determine the amount of heat in calories that we require to raise the temperature of one gram of a material by one degree Celsius.

In simple words, the specific heat equation or the specific heat formula is the ratio of the amount of heat required to raise the temperature of a substance by one degree celsius to the amount of heat required to raise the temperature of the equal amount of water by one-degree Celsius at the room temperature.

The term specific heat is generally used when we are referring to something particular, such as we are referring to raising the temperature of a particular substance. Now, usually, everyone gets confused between the specific heat and the heat capacity. 

The heat capacity is the ratio of the quantity of heat required to alter the temperature by one degree Celsius, whereas the specific heat is the amount of heat required to raise the temperature of the substance, not just to change or alter the temperature of the material. One important point to note down here is that when we consider a particular amount of mass we tend to use the word Specific Heat or Specific Heat Capacity. The specific heat capacity of most of the thermodynamic systems is not constant and it depends on physical entities such as pressure, volume, and temperature.

Derivation

Now, let us have a look at the specific heat formula, which is denoted by the letter C. According to the definition of the specific heat, we have seen that it is directly proportional to the temperature change (in particular raising temperature). We know that amount of heat required (Q) is directly proportional to the change in temperature, thus we write:

⇒ Q ∝ ΔT ……(1)

Where,

ΔT is the change or raise in the temperature

Also, we know that the specific heat is also directly proportional to the mass of the substance under consideration, thus we get:

⇒ Q ∝ m ……..(2)

Where,

m is the mass of the substance

Combining equations (1) and (2) we get:

⇒ Q ∝ mΔT

⇒ Q = CmΔT

Where,

C is the specific heat 

Thus, the specific heat equation is given by:

C =  [frac{Q}{m Delta t}]

 

or

C = [frac{Delta Q}{m Delta t}] joules/kg [^{0}]c ……..(3)

 

Where,

ΔQ is the amount of heat gained or lost in joules

ΔT is the change or raise in the temperature in degree celsius

m is the mass of the substance in kg 

Equation (3) is known as the specific heat equation or specific heat formula in physics and it is used to solve specific heat examples and to estimate specific heat of substance in real life. The specific heat substance will vary depending on the type of substance.

Specific Heat Examples

Now, let us have a look at a few specific heat examples that will help us to understand the concept of specific heat and the specific heat equation or the specific heat formula in a better way. 

1) If 968 joules of heat is required to raise the temperature of the 50 g of substance from 3000c to 4000c. Then, calculate the specific heat of the substance.

Solution:

Given, amount of heat required = Q = 960 joules

The mass of the substance = m = 50 g

The change in temperature = ΔT = T₂ – T₁ = 40 – 30 = 100C 

Now, we are asked to determine the specific heat of the substance. We know that the specific heat or the specific heat capacity is given by the equation:

⇒  C = [frac{Q}{m Delta T}]

Where,

Q – The amount of heat required

ΔT – The change or raise in the temperature

m – The mass of the substance

Substituting the required values in the above equation we get:

⇒ C = [frac{Q}{m Delta T}] = [frac{968}{50 times 10}]

 = 96850×10

⇒ C = 1.936 joules/g 0c

Therefore, the specific heat capacity of the substance is 1.94 joules/g 0c.

2) Calculate the heat required to raise 6kg of water from 400 C to 800 C? (Specific heat of water = 4.2 X 10³ J/kg 0c)

Solution:

Given, the mass of the water = m = 6kg

The change in temperature = ΔT = T₂ – T₁ = 80 – 40 = 400c  

The specific heat capacity of the water = C = 4.2 x 10³J/kg0c 

Now, our aim is to determine the amount of heat required to raise the temperature of 6kg of water from 4000c to 8000c. Thus, we know that the amount required is given by the equation:

⇒ Q = CmΔT

Where,

C – The specific heat of the substance

ΔT – The change or raise in the temperature

m – The mass of the substance

Substituting, required values in the above expression:

⇒ Q = CmΔT = (4.2 x 10³) (6) (40)

⇒ Q = 10.08 x 10⁵ J

⇒ Q = 1008 kJ

Therefore, the amount of heat required to raise 6kg of water from 4000c to 8000c is 1008 kJ.

Three Different Formula of Initial Velocity Derived from Equations of Motion

Motion is an important part of the concepts you learn in physics. It describes how a body moves from one location to the other at a particular time period. This is where you will also learn the three laws of motion. A new term will emerge with the name initial velocity. In this section, we will discuss and learn the meaning and use of the initial velocity formula in different aspects. We will also study how this formula is being derived and used to solve problems.

Initial Velocity: What is it and How Its Formula is Derived?

In the advanced classes of physics, we learn how to determine the value of a body’s motion. This becomes difficult when the body is not in uniform motion. The motion changes with time. It either increases or decreases. This is when the concept of initial velocity comes into the discussion. It is where we will derive and use the initial vertical velocity formula.

Let us suppose the initial velocity is denoted by ‘u’, the final velocity is denoted by ‘v’, the time taken to increase the velocity from ‘u’ to ‘v’ is‘t’, then the acceleration ‘a’ is:

a = (v-u)/t

If we rearrange it, we will find the formula for the first equation of motion. It is:

v = u + at

Now, if we rearrange it again, we will find the formula of the initial velocity equation to determine its value.

u = v – at

If we do the same rearrangement to all the equations of motion, we will find three such unique initial horizontal velocity formula to identify the value of the initial velocity. Let us take a quick look at those formulas.

The first formula to find initial velocity is u = v – at.

The Second Formula is:

u = s/t – ½ at where ‘s’ = displacement travelled within the time period ‘t’

The third initial velocity formula physics is derived from the 3rd equation. It is:

u2 = v2 – 2as

Why Should We Learn Three Different Initial Velocity Formula?

The prime reason for learning three different formulas for identifying the initial velocity is convenience. You can see that all the formulas have something unique. The first formula can derive the value of initial velocity without ‘s’ or displacement used. The second formula can derive the value of ‘u’ without ‘v’. Similarly, the third formula can derive the value of ‘u’ without using ‘t’.

Conclusion

Learning three different formulas with the initial velocity equation will help you identify the value of initial velocity easily. You will also understand the equations of motion better and use them to calculate the values of any term using the data given in a problem.