**This article is formulated according to the 4th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.**

When a body moves in a plane (a two-dimensional motion) or in a space (a three-dimensional motion) then the position, displacement, velocity and acceleration of the body have two or three components respectively. Then we need to use Vectors to describe the concept of position, displacement, velocity and acceleration.

## Scalar quantity

A physical quantity having only magnitude is called a scalar quantity. It is specified completely by a single number along with proper unit. Ex: mass, length, temperature, speed, charge, area etc. Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers. They follow the rules of algebra.

## Vector quantity

A Physical quantity having both magnitude and direction and obey the triangle law of addition is called Vector quantity. It is represented by a number with an appropriate unit and direction. Ex: Displacement, velocity, acceleration, force, momentum etc.

## Differences between Scalar quantity and Vector quantity

Scalar Quantity | Vector Quantity |

It has only magnitude | It has both magnitude and direction |

They follow the rules of ordinary algebra | They follow the rules of vector algebra |

These change when magnitude changes | These changes when magnitude changes or direction changes or both of them changes. |

Ex: Mass, Length, Temperature, Area | Ex: Displacement, velocity, Acceleration, Force |

## Representation of a vector

To represent a vector we use a bold face letters or an arrow placed over a letter.

Ex:

Here O is called the initial point and P is called the terminal point. The length of the line segment OP represents the magnitude and the arrow at the endpoint indicates the direction. The magnitude of a vector is often called the absolute value and indicated by,

## Classification of vectors

## Parallel vectors

Two or more vectors having same direction are called parallel vectors.

## Anti-parallel vectors (opposite vectors)

Vectors having opposite directions are called anti-parallel vectors (opposite vectors).

## Equality of vector (Equal vectors)

Two (or more) vectors having same magnitude and direction, representing the same physical quantity are called Equal vectors.

## Negative of a vector

A vector having same magnitude but having opposite direction to that of the given vector is called negative of a given vector.

## Zero(Null) vector

A vector whose magnitude is zero is called Zero vector. It is represented by 0ββ and the direction is not specified.

Properties of Zero vector are, π΄β β π΄β = 0ββ

|0ββ| = 0

π΄β β 0ββ = π΄β

0ββ π΄β = π΄β 0ββ = 0ββ

π 0ββ = 0ββ

## Unit vector

A vector having unit magnitude is called unit vector. Its purpose is to specify a direction. Unit vector has no dimensions and unit. If πβ is a vector, then the unit vector in direction of πβ is written as πΜ (read as βa capβ) πβ = |πβ| πΜ, Then, mathematically unit vector can be represented as, πΜ = πβ |πβ|

Note: The unit vectors in the positive directions of x, y and z axes are labeled as π,Μ πΜ πππ πΜ respectively.

## Addition of vectors β Graphical method

Two vectors representing the same quantity in the same unit are added using the following rules.

**(i) Triangle method of vector addition: Law of the triangle of vectors or Triangular law of vector addition:**

If two vectors πβ and πββ are represented by two sides of a triangle in head-to-tail form, then the closing side of the triangle taken from the tail of the first to head of the second represents the vector sum of πβ and πβ.

**Explanation: **Consider two vectors, πβ = π΄π΅β and πββ = π΅πΆβ are of same nature. The triangle ABC is completed by joining A and C. According to the triangle law of addition, π΄πΆβ = πβ represents the sum of πβ and πβ. πβ + πβ = πβ or π΄π΅β + π΅πΆβ = π΄πΆβ

Note: In this procedure of vector addition, vectors are arranged head-to-tail. Hence it is called the head-to-tail method.

## Properties vector addition

(a) Vector addition is commutative.

(b) Vector addition is Associative.

πβ + πβ = πβ = πβ + πβ (πβ + π) + πβ = πβ + (πβ + πβ)

(ii) Parallelogram method of vector addition

## Law of parallelogram of vectors or Parallelogram law of vector addition

If two vectors are represented by two adjacent sides of a parallelogram, then the diagonal drawn from the common initial point represents their vector sum.

**Explanation:** Vector πβ and πβ are drawn with a common initial point and parallelogram is constructed using these two vectors as two adjacent sides of a parallelogram. The diagonal originating from the common initial point is vector sum of πβ and πβ.

## Subtraction vectors β Graphical method

Subtraction of vectors can be defined in terms of addition of vectors. Consider two vectors πβ and πβ of same nature and another vector βπβ which is opposite (negative) vector of πβ , then πβ + ( βπβ) = πβ + πβ

**Note:** Subtraction vector is neither commutative nor associative.

## Multiplication of a vector by real (Scalar) number OR Scalar multiplication of a Vector

The product of a vector π£β and a positive number (Scalar) π gives a vector, whose magnitude is changed by a factor π but direction is same as that of π£β. |ππ£β| = π|π£β | (ππ π > 0) If π is negative, the direction of the vector ππ£β is opposite to the direction of the vector π£β and magnitude is β π times |π£β |. If the multiplying factor π is dimensionless then ππ£β have the same dimensions as that of π£β and is product of dimensions if π has dimensions.

## Resolution of vectors

Splitting a given vector into a number of components is called resolution of vectors OR The process of finding the components of a given vector is called resolution the vector.

## Expressions for X and Y components of a Vector

Consider a vector ππ΄β = πβ in X-Y plane, which makes an angle π with the positive X-axis. Draw AM and AN perpendicular to X and Y axes respectively. Let ππβ = πβ_{π₯} and ππβ = πβ_{π¦}. From parallelogram law of addition, we have ππβ + ππβ = ππ΄ββ

πβ_{π₯} + πβ_{π¦} = πβ (Here πβ_{π₯} is x – component of πβ and πβ_{π¦} is y – component of πβ) From βππ΄π, πππ π = ππ/ππ΄ = ππ₯/π

π_{π} = π ππ¨π¬ π½

and π ππ π = π΄π/ππ΄ = π_{π¦}/π

π_{π} = π π¬π’π§ π½

Vectoricaly πβ_{π} = πβ ππ¨π¬ π½ and πβ_{π} = πβπ¬π’π§ π½

**Note:** πβ_{π₯} and πβ_{π¦} being perpendicular are called rectangular components of πβ.

Magnitude

Magnitude of πβ is given by |πβ| = π

Now, π_{π₯}^{2} = π^{2}cos^{2}π and π_{π¦}^{2} = π^{2}sin^{2}π

Taking π_{π₯}^{2} + π_{π¦}^{2} = π^{2}cos^{2}π + π^{2}sin^{2}π

π_{π₯}^{2} + π_{π¦}^{2} = π^{2}(cos^{2}π + sin^{2}π)

π_{π₯}^{2} + π_{π¦}^{2} = π^{2}

π = β(π_{π}^{π} + π_{π}^{π})

ππ’π«ππππ’π¨π§

By taking π_{π¦}_{/}π_{π₯} = πsinπ/πcosπ

π_{π¦}/π_{π₯} = tanπ

π½ = πππ§^{βπ}(π_{π}/π_{π})

**Note:** (i) In terms of unit vectors, πβ = π_{π₯}πΜ+ π_{π¦}πΜ= πcosππΜ + sinππΜ

where πβ_{π₯} = π cos π πΜ

and πβ_{π¦} = π sin π πΜ

(ii) If πβ is in XYZ plane and makes an angle πΌ, π½ and πΎ with X, Y and Z axes respectively, then π_{π₯} = π cos πΌ, π_{π¦} = π cosπ½, π_{π§} = π cos πΎ

and πβ = π_{π₯}πΜ+ π_{π¦}πΜ+ π_{π§}πΜ

The magnitude of πβ is, π = β(π_{π₯}^{2} + π_{π¦}^{2} + π_{π§}^{2}).

Find the magnitude and direction of the resultant of two vectors π¨β and π©β in terms of their magnitudes and angle π½ between them.

Let ππβ and ππβ represent the two vectors π΄β and π΅β making an angle π. Then using the parallelogram method of vector addition ππ β represents the resultant vector π β.

π β = π΄β + π΅β

Draw SN is normal to OP extended. In βπππ, πππ π = ππ/ππ

ππ = ππ cos π = π΅ cos π and π ππ π = ππ/ππ

ππ = ππ sin π = π΅ sin π

Magnitude

From geometry, ππ^{2} = ππ^{2} + ππ^{2}

ππ^{2} = (ππ + ππ)^{2} + ππ^{2}

π
^{2} = (π΄ + π΅ cos π)^{2} + (π΅ sin π)^{2}

π
^{2} = π΄^{2} + (π΅ cos π)^{2} + 2π΄π΅ cos π + π΅^{2} sin^{2}π

π
^{2 }= π΄^{2} + π΅^{2} cos^{2}π + 2π΄π΅ cos π + π΅^{2} sin^{2} π

π
^{2} = π΄^{2} + π΅^{2} (cos^{2}π + sin^{2} π) + 2π΄π΅cosπ

π
^{2} = π΄^{2} + π΅^{2} + 2π΄π΅cos π

πΉ = β(π¨^{π} + π©^{π} + ππ¨π© ππ¨π¬π½)

**Direction:** Let πΌ be angle made by the resultant vector π
β with the vector π΄β, then tan πΌ = ππ/ππ = ππ/(ππ + ππ)

tan πΌ = π΅sinπ/(π΄ + π΅cosπ)

πΆ = πππ§^{β}^{π}(π©π¬π’π§π½/(π¨ + π©ππ¨π¬π½)

## Limitations of Graphical method of adding vectors

(i) It is very difficult method.

(ii) It has limited accuracy.

To overcome these limitations Analytical method of addition of vectors is preferred.

## Addition of vectors β Analytical method

In two Dimensions

Consider two vectors πβ and πβ in X-Y plane. If πβ = π_{π₯}πΜ+ π_{π¦}π Μ and πβ = π_{π₯}πΜ+ π_{π¦}π Μ then, π
β = πβ + πβ

π
β = (π_{π₯}πΜ+ π_{π¦}πΜ) + (π_{π₯}πΜ+ π_{π¦}πΜ)

π
β = π_{π₯}πΜ+ π_{π₯}πΜ+ π_{π¦}πΜ+ π_{π¦}π Μ

π
β = (π_{π₯} + π_{π₯} )πΜ+ (π_{π¦} + π_{π¦})πΜ

πΉβ = πΉ_{π}πΜ+ πΉ_{π}π Μ

Where π
_{π₯} = π_{π₯} + π_{π₯} and π
_{π¦} = π_{π¦} + π_{π¦}

In three Dimensions

If πβ = π_{π₯}πΜ+ π_{π¦}πΜ+ π_{π§}πΜ and πβ = π_{π₯}πΜ+ π_{π¦}πΜ+ π_{π§}πΜ

then, π β = πβ + πβ

π
β = (π_{π₯}πΜ+ π_{π¦}πΜ+ π_{π§}πΜ) + (π_{π₯}πΜ+ π_{π¦}πΜ+ π_{π§}πΜ)

π
β = π_{π₯}πΜ + π_{π₯}πΜ + π_{π¦}πΜ + π_{π¦}πΜ + π_{π§}πΜ + π_{π§}πΜ

π
β = (π_{π₯} + π_{π₯}) πΜ+ (π_{π¦} + π_{π¦}) πΜ+ (π_{π§} + π_{π§}) πΜ

πΉβ = πΉ_{π}πΜ+ πΉ_{π}πΜ+ πΉ_{π}πΜ

This method can be extended to addition and subtraction of any number of vectors.

## Motion in a plane

## Position vector

The position vector πβ of a particle located in X-Y plane with reference to the origin is given by, πβ = π₯πΜ+ π¦πΜ

Where π₯ and π¦ are component of πβ along X-axis and Y-axis respectively.

## Displacement

Consider a particle moves along curve. Initially it is at π_{1} at time π‘_{1 }and moves to a new position π_{2} at time π‘_{2}. Then the displacement is given by, βπβ = πβ_{2} β πβ_{1}

βπβ = (π₯_{2}πΜ+ π¦_{2}πΜ) β (π₯_{1}πΜ+ π¦_{1}πΜ)

βπβ = (π₯_{2}πΜβπ₯_{1}π)Μ + (π¦_{2}_{π}_{Μ}β π¦_{1}πΜ)

βπβ = (π₯_{2}βπ₯_{1}) πΜ + (π¦_{2} β π¦_{1}) πΜ

βπβ = βππΜ+ βππΜ

Where βπ₯ = π₯_{2} β π₯_{1} and βπ¦ = π¦_{2} β π¦_{1}

## Velocity

## Average velocity

It is defined as ratio of the displacement to the time taken.

π£βΜ = βπβ/βπ‘ = (βπ₯πΜ+ βπ¦πΜ)/βπ‘

π£βΜ = (βπ₯/βπ‘) πΜ+ (βπ¦/βπ‘) πΜ

πβΜ
= πΜ
_{π}πΜ+ πΜ
_{π}πΜ

Where πΜ
_{π} = βπ₯/βπ‘ and πΜ
_{π} = βπ¦/βπ‘

Direction of the average velocity is same as that of the displacement.

## Instantaneous velocity (Velocity)

It is given by the limiting value of the average velocity as the time interval approaches to zero.

π£β = lim _{β}_{π‘}_{β0} βπβ/βπ‘

πβ = π πβ/π π

The direction of velocity at any point on the path of the object is tangential to the path at that point and in the direction of the motion. The components of the velocity π£β are given by, π£β = lim _{β}_{π‘}_{β0} βπβ/βπ‘

π£β = lim _{β}_{π‘}_{β0} (βπ₯/βπ‘) πΜ+ (βπ¦/βπ‘) πΜ

π£β = lim _{β}_{π‘}_{β0} (βπ₯/βπ‘) πΜ + lim _{β}_{π‘}_{β0} (βπ¦/βπ‘) πΜ

π£β = (ππ₯/ππ‘) πΜ+ (ππ¦/ππ‘) πΜ

πβ = π_{π}πΜ+ π_{π}πΜ

The magnitude is given by π = β(π_{π}^{π} + π_{π}^{π}) and Direction is given by

π½ = πππ§^{βπ}(π_{π}/π_{π})

## Acceleration

## Average acceleration

It is defined as the change in velocity divided by time interval.

πΜ
β = βπ£β/βπ‘ = β(π£_{π₯}πΜ+π£_{π¦}πΜ)/βπ‘

πΜ
β = (βπ£_{π₯}/βπ‘) πΜ + (βπ£_{π¦}/βπ‘) πΜ

πβΜ
β = πΜ
_{π}πΜ+ πΜ
_{π}πΜ

## Instantaneous acceleration (Acceleration)

It is the limiting value of the average acceleration as the time interval approaches zero.

πβ = lim_{β}_{π‘}_{β0} (βπ£β/βπ‘)

πβ = π πβ/π π

The Components are given by, πβ = lim _{β}_{π‘}_{β0} (βπ£β/βπ‘)

πβ = lim _{β}_{π‘}_{β0} β(π£_{π₯}πΜ+ π£_{π¦}πΜ)/βπ‘

πβ = lim _{βπ‘β0} (βπ£_{π₯}/βπ‘) πΜ+ lim _{βπ‘β0} (βπ£_{π¦}/βπ‘) πΜ

πβ = (ππ£_{π₯}/ππ‘) πΜ+ (ππ£_{π¦}/ππ‘) πΜ

πβ = π_{π}πΜ+ π_{π}πΜ

In one dimension the direction of velocity and acceleration is same or in opposite direction but in two or three dimensions, velocity and acceleration vectors may have any angle between 0Β° and 180Β°.

## Motion in a plane with constant acceleration

Consider an object moving in X-Y plane and its acceleration πβ is constant. Let the velocity of the object be π£β_{0} at time π‘ = 0 and π£β at time π‘, then

(i) π£β = π£β_{0} + πβπ‘

In terms of its components,

π£_{π₯} = π£_{0}_{π₯} + π_{π₯}π‘

π£_{π¦} = π£_{0}_{π¦} + π_{π¦}π‘

(ii) Displacement is πβ β πβ_{0} = π£β_{0}π‘ + (1/2)πβπ‘^{2}

In terms of its components,

π₯ β π₯_{0} = π£_{0}_{π₯}π‘ + (1/2)π_{π₯}π‘^{2}

π¦ β π¦_{0} = π£_{0}_{π¦}π‘ + (1/2)π_{π¦}π‘^{2}

(iii) π£ β^{2} = π£_{0} β^{2} + 2πβ(πβ β πβ_{0})

In terms of its components,

π£_{π₯}^{2} = π£_{0}_{π₯}^{2} + 2ππ₯(π₯ β π₯_{0})

π£_{π¦}^{2} = π£_{0}_{π¦}^{2} + 2ππ¦(π¦ β π¦_{0})

The motion in plane can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.

## Relative velocity in two Dimensions

Suppose two objects π΄ and π΅ are moving with velocities π£β_{π΄} and π£β_{π΅}, then the velocity of the object π΄ relative to that of π΅ is, π£β_{π΄π΅} = π£β_{π΄} β π£β_{π΅}.

Similarly, the velocity of the object π΅ relative to that of π΄ is, π£β_{π΅π΄} = π£β_{π΅} β π£β_{π΄} Therefore, π£β_{π΄π΅} = βπ£β_{π΅π΄} and |π£β_{π΄π΅}| = |π£β_{π΅π΄}|

## Examples for Motion in a plane

**(i) Projectile motion (Uniformly accelerated motion)**

**(ii) Circular motion (non-uniformly accelerated motion)**

When a particle traces a curve in two-dimensional plane, the velocity of the particle changes at least in direction. Hence, a two-dimensional motion along a curve is essentially an accelerated motion. Acceleration may be uniform or non-uniform.

## Projectile

A projectile is any object thrown into air or space.

## Projectile motion

Motion associated with a projectile in parabolic path is called Projectile motion. Ex: A ball leaving the hand of a bowler, A stone thrown at an angle to the horizontal, an object dropped from an aeroplane in horizontal flight. The motion of projectile may be thought of as the result of two separate, simultaneously occurring components of motion. One component is along a horizontal direction with any acceleration and other along the vertical direction with constant acceleration due to gravity. It was Galileo, who first stated this independency of the horizontal and vertical components of projectile motion.

## Analysis of Projectile motion

Let a projectile is projected with initial velocity π£β_{0} that makes an angle π with x-axis. The acceleration acting on it is due to gravity and is directed vertically downwards. π_{π₯} = 0, π_{π¦} = βπ, hence πβ = βππΜ.

The components of initial velocity π£β_{0} are, π£_{0}_{π₯} = π£_{0}cosπ, π£_{0}_{π¦} = π£_{0}sinπ

The components of velocity at time π‘ are, π£_{π₯} = π£_{0} cosπ, π£_{π¦} = π£_{0} sin π β ππ‘.

The components of displacements at time π‘ are,

(i) π₯ = π£_{0}_{π₯}π‘ + (1/2)π_{π₯}π‘^{2}

π₯ = π£_{0}_{π₯}π‘ (β΅ π_{π₯} = 0), π = (π_{π} ππ¨π¬ π½) π (along X-axis)

(ii) π¦ = π£_{0}_{π¦}π‘ + (1/2)π_{π¦}π‘^{2}, π = (π_{π} π¬π’π§π½)π β (π/π)ππ^{π} (along π β axis)

## Path of a projectile

The path described by the projectile is called trajectory. The trajectory is a parabola.

## Expression for path of a projectile (Show that the path of a projectile is Parabola)

The displacement of the projectile along X-axis is, π₯ = (π£_{0} cos π)π‘, π‘ = π₯/π£_{0}cos π The displacement of the projectile along Y-axis is,

π¦ = (π£_{0} sin π)π‘ β (1/2)ππ‘^{2}

π¦ = (π£_{0} sin π) (π₯/π£_{0} cos π) β (1/2)π(π₯/π£_{0}cos π)^{2}

π¦ = (sin π/cos π)π₯ β (1/2)(π/π£_{0}^{2} cos^{2}π)π₯^{2}

π¦ = π₯ tan π β (1/2)(π/π£_{0}^{2}cos^{2}π)π₯^{2}

π = π_{π} β π_{π}^{π}

where π = π‘πππ and π = (1/2)ππ£_{0}^{2}πππ ^{2}π

The equation π¦ = ππ₯ β ππ₯^{2} represents a parabola.

Hence the trajectory is a parabola.

## Time of Flight

It is the time during which the projectile is in flight. It is denoted by π_{π}.

## Expression for Time of flight

The component of velocity along Y-axis at time t is, π£_{π¦} = π£_{0 }sinπ β ππ‘

At maximum height π£_{π¦} = 0 and time for maximum height, π‘ = π‘_{π}.

0 = π£_{0} sin π β ππ‘_{π}

ππ‘_{π} = π£_{0} sinπ

π‘_{π} = π£_{0} sinπ/π

Time of flight π_{π} = 2π‘_{π} because βtime of ascent = time of descentβ

π»_{π} = ππ_{π}π¬π’π§π½/π

## Maximum height of a projectile

It is the maximum height reached by the projectile in time π‘_{π}. It is denoted by β_{π}.

## Expression for maximum height of a projectile

The displacement along Y β axis is, π¦ = (π£_{0} sinπ)π‘_{π} β (1/2)ππ‘_{π}^{2}

β_{π} = (π£_{0} sin π)(π£_{0} sin π/π) β (1/2)π(π£_{0}sin π/π)^{2}

β_{π} = (π£_{0} sin π)^{2}/π β (1/2)π((π£_{0} sinπ)^{2}/π^{2})

β_{π} = (π£_{0} sin π)^{2}/π β (1/2)(π£_{0} sinπ)^{2}π

β_{π} = (1 β (1/2))(π£_{0} sinπ)^{2}/π

β_{π} = (1/2)(π£_{0} sinπ)^{2}/π

π_{π} = (π_{π}π¬π’π§π½)^{π}/ππ

## Horizontal Range of projectile

It is the horizontal distance covered by the projectile during its flight. It is denoted by π .

## Expression for Horizontal Range of projectile

Displacement along X-axis is, π₯ = (π£_{0 }cosπ)π‘

Now π₯ = π
and π‘ = π_{π}

π
= (π£_{0} cosπ)π_{π}

π
= (π£_{0} cosπ)(2π£_{0} sinπ/π)

π
= (π£_{0}^{2}/π)(2cosπsinπ)

πΉ = (π_{π}^{π}/π)π¬π’π§ππ½

**Note:**

(i) For a given speed of projection, the projectile will have a maximum range (π π) when sin 2π is maximum or the angle of projection is 45Β°. sin 2π = 1 βΉ 2π = 90Β° Then angle of projection, π = 45Β°.

Maximum range, π
_{π} = (π£_{0}^{2}/π) sin2(45Β°)

πΉ_{π} = π_{π}^{π}/π

(ii) Show that πΉ_{π} = ππ_{π} when the angle of projection is π½ = ππΒ°.

For π = 45Β°, π
_{π} = π£_{0}^{2}/π and β_{π} = π£_{0}^{2}sin^{2}π/2π = π£_{0}^{2}sin^{2}45Β°/2π = (π£_{0}^{2}/2π) Γ (1/2) β_{π} = π£_{0}^{2}/4π = (1/4)(π£_{0}^{2}/π)

β_{π} = (1/4)(π
_{π})

πΉ_{π} = ππ_{π}

Since the maximum range of a projectile is equal to 4 times the maximum height reached.

## Uniform circular motion

The motion of the object in a circular path at a constant speed is called uniform circular motion. Even though the object moves at a constant speed it has acceleration, because there is a continuous change in its direction of motion. Hence there is a change in its velocity from point to point.

## Expression for Acceleration

Let πβ and πββ² be the position vectors and π£β and π£ββ² are the velocities of the object when it is at π and π as shown. Velocity at a point is along the tangent at that point in the direction of motion. From βπ΄π΅πΆ, π΄π΅β + π΅πΆβ = π΄πΆβ

π£β + π΅πΆβ = π£ββ²

π΅πΆβ = π£ββ² β π£β = βπ£β

βπ£β is the change in velocity, which is towards the center. Since the path is circular, π£β and π£ββ² are perpendicular to πβ and πββ² respectively. Therefore βπ£β is perpendicular to βπβ. The average acceleration = βπ£β/βπ‘.

Since βπ£β is perpendicular to βπβ, πΜ is along βπ£β and perpendicular to βπβ and directed towards the center of the circle. The Instantaneous acceleration is,

πβ = lim _{β}_{π‘}_{β0} βπ£β/βπ‘

Its magnitude is given by,

π_{π} = |πβ|

π_{π} = lim _{β}_{π‘}_{β0} |βπ£β|/βπ‘

Since the velocity vectors π£β and π£ββ² are always perpendicular to πβ and πββ² , the angle between π£β and π£ββ² is also βπ. Since βπ΄π΅πΆ and βπππ are similar.

Then, π΅πΆ/ππ = π΄π΅/ππ

|βπ£β|/|βπβ| = π£/π (β΅ |πβ| = π = πππππ’π )

βπ£β = (|βπβ|π£)/π

β΄ π_{π} = lim _{β}_{π‘}_{β0} (|βπβ|π£/βπ‘π
)

π_{π} = (π£/π
)lim _{β}_{π‘}_{β0} |βπβ|/βπ‘

π_{π} = (π£/π
)π£

π_{π} = π^{π}/πΉ

This equation represents the magnitude of acceleration and is directed toward the center.

## Centripetal acceleration

The acceleration, which is directed towards the center, is called centripetal acceleration. The term centripetal acceleration was termed by Newton and Centripetal comes from a Greek term that means Centre seeking towards the center.

**Note:**

(i) In uniform circular motion as the object moves from P to Q (in the above figure) in time βπ‘ the line OP turns through an angle βπ, called angular distance. But angular speed, π = βπ/βπ‘ If the distance traveled ππ = βπ then, Speed, π£ = βπ /βπ‘ But βπ = π βπ, where R is the radius of the trajectory.

β΄ π£ = π βπ/βπ‘ = π (βπ/βπ‘)

π = πΉπ β (1)

The Centripetal acceleration π_{π} = π£^{2}/π
= π
^{2}π^{2}/π

π_{π} = πΉπ^{π}

(ii) Time period (T)

Time is taken by an object to make one revolution.

(iii)Frequency (π)

The number of revolutions made in one second. π = 1/π.

Distance moved in time period π = 2ππ

Speed, π£ = 2ππ /π = 2ππ π

π£ = π (2ππ) β (2)

Comparing equations (1) and (2), we get π = ππ π

Then Acceleration, π_{π} = (2ππ)^{2}π

π_{π} = ππ
^{π}π^{π}πΉ

## Problems

1) A cricket ball is thrown at a speed of 28ππ ^{β1} in a direction of 30Β° with the horizontal.

Calculate

(a) the maximum height

(b) the time taken by the ball to return to the same level.

(c) the distance from the thrower to the point where the ball returns to the same level.

Given π£_{0} = 28ππ ^{β1}, π = 30Β°

(a) β_{π} = (π£_{0} sin π)^{2}/2π

β_{π} = (28 Γ sin 30Β°)^{2}/(2 Γ 9.8)

β_{π} = (28 Γ (1/2)^{2}/(2 Γ 9.8)

β_{π} = 14^{2}/19.6

β_{π} = 10π

(b) π_{π} = 2π£_{0} sin π/π

π_{π} = (2 Γ 28 Γ sin 30Β°)/9.8 = (2 Γ 28 Γ (1/2))/9.8

π_{π} = 2.85π

(c) π
= (π£_{0}^{2}/π)sin2π

π
= (28^{2}/9.8)sin(2 Γ 30Β°) = (28^{2}/9.8) Γ 0.866

π = 69.28π

2) A cricket ball projected at an angle of 30Β° with the horizontal takes 3 seconds to reach the ground.

Calculate

(a) the velocity of the projection.

(b) The horizontal range of the ball.

Given, = 30Β°, π_{π} = 3π

(π) π_{π} = 2π£_{0} sin π/π

3 = (2 Γ π£_{0} Γ sin 30Β°/9.8)

π£_{0} = (3 Γ 9.8)/(2 Γ sin 30Β°) = (3 Γ 9.8)/(2 Γ (1/2))

π£_{0} = 29.4ππ ^{β1}

(π) π
= (π£_{0}^{2}/π)sin2π

π
= ((29.4)^{2}/9.8) sin(2 Γ 30Β°) = (864.36/9.8) Γ 0.866

π = 76.38π

3) The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40ππ ^{β1} can go without hitting the ceiling of the hall?

Given π£_{0} = 40ππ ^{β1} , Height of the hall, β_{π} = 25π, π = ?, β_{π} = (π£_{0} sin π)^{2}/2π

25 = (40 Γ sin π)^{2}/(2 Γ 9.8)

25 Γ 2 Γ 9.8 = (40 Γ sin π)^{2}

40 Γ sin π = β(25 Γ 2 Γ 9.8)

sin π = β(25 Γ 2 Γ 9.8)/40 = 0.5534

π = sin^{β1}(0.5534) = 33.6Β°

π
= (π£_{0}^{2}/π)sin 2π

π
= (40^{2}/9.8) sin(2 Γ 33.6Β°)

π = (1600/9.8) Γ 0.9219

π = 150.5π

4) A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground can the cricketer throw the same ball?

Given π
_{π} = 100π,

For maximum range, π = 45Β°

π
_{π} = π£_{0}^{2}/π

100 = π£_{0}^{2}/9.8

π£_{0}^{2} = 100 Γ 9.8 = 980

π£_{0} = 31.3ππ ^{β1}

To throw the ball vertically upwards, π = 90Β°

β_{π} = (π£_{0} sin π)^{2}/2π

β_{π} = (31.3 Γ sin 90Β°)^{2}/(2 Γ 9.8)

β_{π} = (31.3 Γ 31.3)(2 Γ 9.8)

β_{π} = 49.98π β 50π

5) A stone tide to the end of a string 80cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25s, what is the magnitude and direction of acceleration of the stone?

Given Radius, π = 80ππ = 0.80π

Frequency, π = 14/25 π ^{β1}

π = 2ππ

π = 2 Γ (22/7) Γ (14/25)

π = 3.52 πππ/π

π_{π} = ππ^{2}

π_{π} = 0.80 Γ (3.52)^{2}

π_{π} = 9.91ππ ^{β2}

6) An aircraft executes a horizontal loop of radius 1km with a speed of 900kmph. Compare its centripetal acceleration with the acceleration due to gravity.

Given Radius, π = 1ππ = 1000π

Speed, π£ = 900πππβ = 900 Γ (1000/3600)ππ ^{β1} = 250ππ ^{β1}

π_{π} = π£^{2}/π

π_{π} = 250^{2}/1000 = 62500/1000

π_{π} = 62.5ππ ^{β2}

Comparison: But π = 9.8ππ ^{β2}

π_{π}_{/}π = 62.5/9.8 = 6.38

π_{π} = 6.38π

7)Rain is falling vertically with a speed of 30ππ ^{β1}. A woman rides a bicycle with a speed of 10ππ ^{β1} in the north-south direction. What is the direction in which she should hold her umbrella?

Given Velocity of bicycle, π£β_{π΅} = 10ππ ^{β1}

Velocity of rain, π£β_{π
} = 30ππ ^{β1}

Magnitude of velocity, π£_{π
π΅} = β(π£_{π΅}^{2} + π£_{π
}^{2})

π£_{π
π΅} = β(30^{2} + 10^{2}) = β1000

π£_{π
π΅} = 31.6ππ ^{β1}

Direction, π = tan^{β1}(π£_{π΅}/π£π
)

π = tan^{β1} (10/30) = tan^{β1}(0.3333)

π = 18Β°26β² She should hold the umbrella at an angle 18Β°26β² with the vertical in south-west direction.

8) An aeroplane flying at 540kmph drops a missile towards the ground. If the height of the plane is 1000m then calculate

(a) time taken by the missile to hit the ground.

(b) Horizontal distance covered by the missile from the initial point.

Given π_{π¦} = 9.8ππ ^{β2}, π_{π₯} = 0, π¦ = 1000π.

At the top, π£_{0}_{π₯} = 540πππβ = 150ππ ^{β1}, π£_{0}_{π¦} = 0

(a) Time taken can be calculated as, π¦ = π£_{0}_{π¦}π‘ + (1/2)π_{π¦}π‘^{2}

1000 = 0 + (1/2) Γ 9.8 Γ π‘^{2}

π‘^{2} = (1000 Γ 2)/9.8

π‘ = 14.29π

(b) π₯ = π£_{0}_{π₯}π‘ + (1/2)π_{π₯}π‘^{2}

π₯ = 150 Γ 14.29

π₯ = 2142.86π

9) Two concurrent forces 20N and 30N are acting at an angle of 60Β° with respect to each other. Calculate the magnitude and direction of the resultant. Given π΄ = 20π, π΅ = 30π, π = 60Β°

(a) Magnitude, π
= β(π΄^{2} + π΅^{2} + 2π΄π΅ cosπ)

π
= β(20^{2} + 30^{2} + (2 Γ 20 Γ 30 Γ cos 60Β°)

π = β(400 + 900 + (2 Γ 600 Γ (1/2)) = β(1900)

π = 43.50π

(b) Direction, πΌ = tan^{β1}(π΅ sinπ/(π΄ + π΅ cos π))

πΌ = tan^{β1}(30 Γ sin 60Β°/(20 + (30 Γ cos 60Β°))

πΌ = tan^{β1}(25.98/35) = 36Β°58β²

The resultant force is 43.50π in the direction of 36Β°58β² with respect to 20π.

## Important Questions.

**One mark.**

1) What is unit vector?

2) What is Zero (null) vector?

3) Is scalar multiplied by a vector, a vector or a scalar?

4) What is the minimum number of vectors to give zero resultant?

5) When will be the resultant of two given vectors is maximum?

6) What is resolution of vector?

7) What is time of flight of a projectile?

8) At what angle range of a projectile is maximum? or when the range of a projectile does become maximum?

9) What is the relation between maximum height and maximum range of a projectile?

10) For angle of projection 30Β°, π is the range of the projectile. Then write another angle of projection for which the range is same.

11) Represent the unit vector in mathematical form.

**Two marks.**

1) What are scalar and vector? Give example. OR distinguish between scalar and vector.

2) What is a projectile? Give an example.

3) Write the equation for the trajectory of a projectile motion. What is the nature of its trajectory?

4) State and explain parallelogram law of vector addition.

5) A unit vector is represented by π΄πΜ+ π΅πΜ+ πΆπΜ. If the value of π΄ and π΅ are 0.5 and 0.8 respectively, then find the value of πΆ.

**Three marks.**

1) State and explain the triangle law of vectors addition.

2) Obtain an expression for maximum height reached by a projectile.

3) Obtain an expression for time of flight of a projectile.

4) What is resolution of vectors? Write expressions for π₯ and π¦ components (Rectangular) of a vector. or obtain the equations for rectangular components of a vector in two dimensions.

5) Derive an expression for magnitude of resultant of two concurrent vectors. or find the magnitude of the resultant of two vectors π΄ and π΅ in terms of their magnitude and angle π between them.

6) Obtain the expression for range of a projectile.

**Five marks**

1) What is centripetal acceleration? Derive an expression for centripetal acceleration of a particle in uniform circular motion. or What is centripetal acceleration? Derive the expression for radial acceleration.

2) What is projectile motion? Show that trajectory of projectile is a parabola. or What is projectile motion? Derive an expression for trajectory of projectile. or show that the path of the projectile is a parabola.

## Additional Problems

1) A bullet is fired at a velocity of 392ππ ^{β1} at an angle of 30Β° to the horizontal. Find the maximum height attended ad time of flight.

2) A body is projected with a velocity of 50ππ ^{β1} in a direction making an angle of 30Β° with the horizontal.

Find

(a) The maximum height.

(b) the time taken by the body to return to the same level

and (c) the range.

3) A ball is thrown into air with a speed of 62ππ ^{β1} at an angle 45Β° with the horizontal.

Calculate

(a) The maximum height attained.

(b) the time of flight.

(c) the horizontal range.

4) A player hits a cricket ball at angle of 40Β° to the horizontal. If the ball moves with a velocity of 20ππ ^{β1}. Find

(a) The maximum height reached by the ball.

(b) the time of flight.

(c) the horizontal range.

Given π = 10ππ ^{β1}.

5) A football player kicks a ball at an angle of 30Β° to the horizontal with an initial velocity of 15ππ ^{β1}. Assuming that the ball travels in a vertical plane. Calculate (a) The maximum height reached. (b) the time of flight and (c) the horizontal range.