Samacheer Class 11 Physics - Kinematics

The different types of motions are:
a) Linear motion: An object is said to be in linear motion if it moves in a straight line.
Example: An athlete running on a straight tack.
b) Circular motion: It is defined as a motion described by an object traveling a circular path.
Example: The motion of a satellite around the earth
C) Rotational motion: If any object moves in a rotational motion about an axis the motion is rotational motion. During rotation, every point in the object traverses a circular path about an axis.
Example: Spiring of earth about its own axis
D) Vibratory motion: If an object or a particle executes to and fro motion about the fixed point it is said to be in vibratory motion. Sometimes called oscillatory motion.
Example: Vibration of a string on a Guitar.

Motion in one dimension: One-dimensional motion is the motion of a particle moving along a straight line.
Example: An object falling freely under gravity close to the earth.
Motion in two dimensions: If a particle is moving along a curved path in-plane, then it is said to be in two-dimensional motion.
Example: Motion of a coin in a carom board.
Motion in three dimensions: A particle moving in usual three-dimensional space has three-dimensional motion.
Example: A bird flying in the sky.

For a stationary observer, the relative velocity of trees and houses is zero. For the observer sitting in the moving train, the relative velocity of houses and trees are negative. So these objects appear to move in the backward direction.

The different types of vectors are:
1. Equal vectors:
Two vectors \(\vec{A}\) & \(\vec{B}\) are said to be equal when they have equal magnitude and same direction and represent the same physical quantity.
(a) Coilinear vectors: Collinear vectors are those which act along the same line. The angle between them can be 0° or 180°
(i) Parallel vectors – If two vectors \(\vec{A}\) & \(\vec{B}\) act in the same direction along the same line or in parallel lines. Angle between them is equal to zero
(ii) Antiparallel vectors:
Two vectors \(\vec{A}\) & \(\vec{B}\) are said to be antiparallel when they are in opposite direction along the same line or in parallel lines. The angle between them is 180°
2. Unit vector:
A vector divided by its own magnitude is a unit vector.
The unit vector of \(\vec{A}\) is represented as \(\hat{A}\)
Its magnitude is equal to 1 or unity
3. Orthogonal unit vectors:
Let \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) be three unit vectors which specify the direction along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directly perpendicular to each other
The angle between any two of them is 90°. Then \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors.

As relative velocity is zero the two bodies A and B have equal velocities. Hence their position-time graphs are parallel straight lines, equally inclined to the time axis.

Yes, the object may be at rest relative to one object and at the same time if maybe in motion relative to another object.
For example, a passenger sitting in a moving train is at rest with respect to his fellow passengers but he is in motion with respect to the objects outside the train. Hence rest and motion are relative terms.

When two objects A and B are moving with uniform velocities then the velocity of one object A with respect to another object B is called the relative velocity of A with respect to B.
Case 1:
Consider two objects A and B moving with uniform velocities \(\overline{V}\) A and \(\overline{V}\) B along straight line in same direction with respect to ground.
The relative velocity of object A with respect to object B is \(\vec{V}\) AB = \(\vec{V}\) A – \(\vec{V}\) B
The relative velocity of object B with respect to object A is \(\vec{V}\) BA = \(\vec{V}\) B –\(\overline{V}\) A
Thus, if two objects are moving it’s the same direction the magnitude of the relative velocity of one object with respect to another is equal to the difference in magnitude of the two velocities.
Case 2:
Consider two objects A and B moving with uniform velocities VA and VB along the same track in the opposite direction
The relative velocity of object A with respect to B is
\(\overline{V}\) AB = \(\overline{V}\) A – ( – \(\overline{V}\) B ) = \(\overline{V}\) A + \(\overline{V}\) B
The relative velocity of object B with respect to A is
\(\vec{V}\) BA = – \(\vec{V}\) B – \(\vec{V}\) A ) = – ( \(\vec{V}\) A + \(\vec{V}\) B )
Thus if two objects are moving in opposite directions the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitudes of their velocities.
Case 3:
Consider two objects A&B moving with velocities VA and VB at an angle 0 between their directions, then the relative velocity of A with respect to B
tan θ = (β is the angle between \(\overline{V}\) AB and VB)
Special cases:
(i) When θ = 0, the bodies move along parallel straight lines in the same direction.
V AB = (V A – V B ) in the direction of V A.
V BA = (V B – V A ) in the direction of V B
(ii) When θ = 180° the bodies move along parallel straight lines in opposite direction.
V AB = V A – (- V B ) = (V A + V B ) in the direction of V A
V BA = ( V B + V A ) in the direction of V B
(iii) If the two bodies are moving at right angles to each other, then θ = 90°
V AB = \(\sqrt{V_{A}^{2}+V_{B}^{2}}\)
(iv) Consider a person moving horizontally with velocity \(\vec{V}\) m Let the rain fall vertically with velocity \(\overline{V}\) R.
An umbrella is held to avoid the rain.
Then relative velocity \(\overline{V}\) M of rain with respect to man is

By definition of velocity v = \(\frac { ds }{ dt }\)
ds = Vdt = (u + at) dt → (1)
when t = (n – 1) second, let distance travelled = S n-1
when t = n, second, let distance travelled = S n

Consider an object thrown horizontally with an initial velocity u, from atop of a tower of height h. The horizontal velocity remains constant throughout its motion and the vertical component of velocity go on increases. The constant acceleration acting along the downward direction is g. The horizontal distance travelled is x(t) = x and the vertical distance travelled is y(t)=y. since the motion is two-dimensional the velocity will have both horizontal (u x ) and vertical (u y ) components.
Motion along horizontal direction:
The particle has zero acceleration along the x-direction and so initial velocity ux remains constant throughout its motion.
The distance travelled by projectile in a time’t’ is given by
x = ut+1/2 at²
x = u x t → (1)
Motion along vertical direction
Here uy =0, a = g, s = y
S = ut + \(\frac { 1 }{ 2 }\) at²
y = \(\frac { 1 }{ 2 }\) gt² → (2)
from (1) t = x/u x sub in equation (2)
y = \(\frac { 1 }{ 2 }\) g (x/ux)²
y = k x² Where k = \(\frac{g}{2 u_{x}^{2}}\).x²
This equation resemble the equation of a parabola. Thus the path followed by the projectile is a parabola.
Expression for time of flight:
The time taken for the projectile to complete its trajectory is called the time of flight.
Let h be the height of the tower or the vertical distance traversed.
Let T be the time of flight w.k. S = ut + 1/2 at²
here s = y = h, u = u y, t = T, a = g
T depends on height of tower or vertical distance & independent of Horizontal velocity.
Expression for Horizontal Range:
The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground it called horizontal range.
w. k. t, S = ut + \(\frac { 1 }{ 2 }\) at²
Here,
t = T, a = 0, S = x = R, u = u x
Hence R ∝ u ∝ & R ∝ \(\frac{1}{\sqrt{g}}\)

The values displayed by Suresh are the presence of mind, helping tendency, and also a sense of social responsibility.
Relative velocity of the bullet with respect to thief’s Jeep = (Vb + Vp)-Vt.
= 180 m/s + 45 km/hr – 155 km/hr
= 180 m/s – 110 x 5/18 m/s
= 180 – 30.5
= 149.5 m/s.

At any instant t, the projectile has velocity components along both the x and y-axis.
The velocity component at any time t along with horizontal component V x = u → (1)
Speed of the projectile when it hits the ground:
When the projectile hits the ground after thrown horizontally from top of tower of height h, the time of flight is T = \(\sqrt{\frac{2 h}{g}}\)
The horizontal component of velocity V x = u
The vertical component of velocity
Conceptual Questions:

No. It is not a projectile. A projectile should have two-component velocities in two mutually perpendicular directions. But in this case, body has a velocity in only one direction.

In a uniform circular motion, the speed of the body remains the same but the direction of motion changes at every point.
Fig. shows the different velocity vectors at different positions of the particle. At each position, the velocity vector V is perpendicular to the radius vector. Thus the velocity of the body changes continuously due to the continuous change in the direction of motion of the body. As the rate of change is of velocity is acceleration a uniform circular motion is an accelerated motion.

If a number of vectors are represented both in magnitude and direction by the sides of an open polygon taken in the same order then their resultant is represented both in magnitude arid direction by the closing side of the polygon taken in the opposite order.

At any given instant of time, the frame of reference with respect to which the position of the object is described in terms of position coordinates (x,y,z) is called “Cartesian coordinate system”.
If x, y, and z axes are drawn in an anticlockwise direction, then the coordinate system is called a right-handed Cartesian coordinate system.

The Scalar product of two vectors (dot product) is defined as the product of the magnitudes of both the vectors and the cosine of angle between them.
If \(\vec{A}\) and \(\vec{B}\) are two vectors having an angle θ between them, then their scalar or dot product is
Example: W = \(\vec{F}\).\(\vec{dr}\). Work done is a scalar product of force \(\vec{F}\) and \(\vec{r}\)

The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, then their vector product is written as \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) which is a vector C defined by \(\overrightarrow{\mathrm{c}}\) = \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = (AB sin 0) \(\hat{n}\)
The direction \(\hat{n}\) of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\), i.e., \(\overrightarrow{\mathrm{c}}\) is perpendicular to the plane containing the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

If the vector product of the two given vectors is having maximum magnitude.
i.e sinθ = 90°, [ (\(\vec{A}\) x \(\vec{B}\))Max = AB\(\hat{n}\) ] then the two vectors are said to be perpendicular.
In this displacement calculator, we will show you how to find displacement in a matter of seconds.

Distance is the actual path length traveled by an object in the given interval of time during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity.

Velocity – Velocity is defined as the rate of change of position vector with respect to time (or) defined as the rate of change of displacement. It Is a vector quantity.
Speed – Speed is defined as the rate of change of distance. It is a scalar quantity.

Acceleration is defined as the rate of change of velocity.
Acceleration \(\vec{a}\) = \(\frac{d \vec{v}}{d t}\)
Acceleration is a vector quantity.
Unit – ms -2
Dimensional formula-[LT -2 ]

• Angular displacement: The angle described by the particle about the axis of rotation in a given time is called angular displacement.
• Angular velocity: The rate of change of angular displacement is called angular velocity.

When an object is moving in a circular path with variable speed, it covers unequal distances in equal intervals of time. Then the motion of the object is said to be a non-uniform circular motion. Here both speed and direction during circular motion change.

The Kinematic equations for angular motion are ω = ω 0 + αt
θ = ω 0 t + \(\frac { 1 }{ 2 }\)αt²
ω² = ω 0 ² + 2αθ
θ = \(\left(\frac{\omega_{0}+\omega}{2}\right)\) x t
ω 0 → initial angular velocity
ω → final angular velocity
α → angular acceleration
θ → angular displacement
t → time interval

In the case of non-uniform circular motion, the particle will have both centripetal and tangential acceleration. The resultant acceleration is obtained as the vector sum of both centripetal and tangential acceleration.
This resultant acceleration makes an angle 6 with a radius vector, which is given by
III. Long Answer Questions:

Let us consider two vector \(\vec{A}\) and \(\vec{B}\) as shown In fig.
Law: To find the resultant of two vectors, the triangular law of addition can be applied as follows.
A and B are represented as the two adjacent sides of a triangle taken in the same order. The resultant is given by the third side of the triangle taken in reverse order.
Magnitude of the resultant vector:
from figure
Let θ be the angle between two vectors.
from ∆ ABN, Sin θ = \(\frac { BN }{ AB }\) ⇒ ∴ BN = B sinθ
Cos θ = \(\frac { AN }{ AB }\) ⇒ ∴ AN = B Cos θ
Which is the magnitude of the resultant \(\vec{A}\) and \(\vec{B}\).
The direction of the resultant vector:
If \(\vec{R}\) makes an angle α with \(\vec{A}\) then

Properties of scalar product:
formula: \(\vec{A}\).\(\vec{B}\) = ABCosθ
1. The product quantity \(\overline{A}\).\(\overline{B}\) is always a scalar. It is positive if the angle between the vectors is acute (θ< 90°) and negative if angle between them is obtuse (90 < θ < 180)
2. The scalar product is commutative \(\overline{A}\).\(\overline{B}\) = \(\overline{B}\).\(\overline{A}\)
3. The scalar product obey distributive law. \(\overline{A}\).( \(\overline{B}\) + \(\overline{C}\) ) = \(\overline{A}\).\(\overline{B}\) + \(\overline{A}\).\(\overline{C}\)
4. The angle between the vector is θ = Cos -1 \(\frac{\bar{A} \cdot \bar{B}}{A B}\)
5. The scalar product of two vectors will be maximum when cos θ = 1 i.e θ = 0 ie when they are parallel.
[ ( \(\overline{A}\).\(\overline{B}\) ) max = AB.]
6. The scalar product of two vectors will be minimum when cos θ = -1 ie θ = 180°
( \(\overline{A}\).\(\overline{B}\)) mm = – AB [the vector are anti-parallel]
7. If two vector \(\overline{A}\) & \(\overline{B}\) are perpendicular to each other then \(\overline{A}\).\(\overline{B}\) = O. Because cos 90 = 0. Then vectors A & B are mutually orthogonal.
8. The scalar product of a vector with it self is termed as self or dot product and is given by
( \(\overline{A}\) )² = \(\overline{A}\).\(\overline{A}\) = AA cos θ = A²
Here 0=0
The magnitude or norm of the vector \(\overline{A}\) is
|A| = A = \(\sqrt{\bar{A} \cdot \bar{A}}\) = A.
9. Incase of orthogonal unit vectors
\(\hat{n}\).\(\hat{n}\) = 1 x 1cos0 = 1
for eg \(\hat{i}\).\(\hat{i}\) = \(\hat{j}\).\(\hat{j}\) = \(\hat{k}\).\(\hat{k}\) = 1
10. Incase of orthogonal unit vectors \(\hat{i}\), \(\hat{f}\), \(\hat{k}\) then \(\hat{i}\).\(\hat{j}\) = \(\hat{j}\).\(\hat{k}\) = \(\hat{k}\).\(\hat{j}\) = 1.1 cos 90 = 0.
11. In terms of components the scalar product of A and B can be written as
Properties of cross product:
Formula \(\vec{A}\) x \(\vec{B}\) = ABsinθ
1. The vector product of any two vectors is always an another vector whose direction perpendicular to the plane containing these two vectors, ie. Orthogonal to \(\overline{A}\) & \(\overline{B}\) even though \(\overline{A}\) & \(\overline{B}\) may not be mutually orthogonal.
2. Vector product is not commutative
\(\overline{A}\) x \(\overline{B}\) = – \(\overline{B}\).\(\overline{A}\)
\(\overline{A}\) x \(\overline{B}\) ≠ \(\vec{B}\) x \(\vec{A}\)
Here magnitude | \(\overline{A}\) x \(\overline{B}\) | = | \(\overline{B}\).\(\overline{A}\) | are equal but opposite direction.
3. The vector product of two vector is maximum when sine = 1, ie θ = 90°
ie. when \(\overline{A}\) and \(\overline{B}\) are orthogonal to each other.
( \(\overline{A}\) x \(\overline{B}\) ) max = AB \(\hat{n}\).
4. The vector product of two non zero vectors is minimum if |sinθ| = 0. ie. θ = 0 or 180°
( \(\overline{A}\) x \(\overline{B}\) ) m in = 0
Vector product of two non zero vectors is equal to zero if they either parallel or anti parallel
5. The self cross product ie product of a vector with itself is a null vector \(\overline{A}\) x \(\overline{B}\) = AA sinθ = 0
6. The self-vector product of the unit vector is zero
i.e. \(\hat{i}\).\(\hat{j}\) = \(\hat{j}\).\(\hat{j}\) = \(\hat{k}\).\(\hat{k}\) = 0
7. In case of orthogonal unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) in accordance with right hand cork screw rule \(\hat{i}\).\(\hat{j}\) =\(\hat{k}\), \(\hat{i}\).\(\hat{k}\) = \(\hat{i}\), \(\hat{k}\).\(\hat{i}\) = \(\hat{j}\) also since cross product is not commutative
9. If two vectors \(\overline{A}\) & \(\overline{B}\) form adjacent sides of a parallelogram then the magnitude of |\(\overline{A}\) x \(\overline{B}\)| will give area 0f parallelogram.
10. Since one can divide a parallelogram into two equal triangles, the area of the triangle is \(\frac { 1 }{ 2 }\) |\(\overline{A}\) x \(\overline{B}\)|.

Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the initial velocity at t = 0, and v be the final velocity after a time of t seconds
(i) Velocity time relation:
The acceleration of the body at any instant is given by first derivative of the velocity with time
a = \(\frac { dv }{ dt }\)
dv = adt
integrating both sides
Displacement time relation:
(ii) The velocity of the body is given by the first derivative of the displacement with respect to time
But v = ds/dt
∴ dv = v dt
v = u + at
ds = (u + at)dt
ds = udt + atdt
Integrating both sides
Velocity-displacement relation:
also we can derive from the relation v = u + at
v – u = at
Substituting in equation s = ut + \(\frac { 1 }{ 2 }\)at²

Consider an object thrown with an initial velocity u at an angle θ with horizontal.
Then initial velocity is resolved into two components
u x = u cos θ horizontally and
u y = u sin θ vertically
At maximum height u y = 0 (since acceleration due to gravity is opposite to the direction of the vertical component).
The Horizontal component of velocity
u x = u cos θ remains constant throughout its motion.
hence after the time t the velocity along the horizontal motion
V x = U x + a x t
= ux = cos θ
The horizontal distance travelled by the projectile in a time ‘t’ is S x = u x t + 1/2 a x t².
Here S x = x u x = u cos θ
a x = 0
∴ x = u cos θt ____ (1)
∴ t = \(\frac { x }{ u cos θ }\) ____ (2)
For vertical motion
V y = u y + a y t
Here v y = v y
u y = u sin θ
a y = – g
v y = u sin θ – gt
The vertical distance travelled by the projectile in the same time ‘t’ is
S y = U y t + a y t²
S y = y, U y = u sin θ a y = – g
y = u sin θ t – 1/2 gt² ____ (4)
Substituting the value of t in (4) we get equation:
Which indicates the path followed by the projectile is an inverted parabola.
Expression for Maximum height:
The maximum vertical distance travelled by the projectile during its motion is called maximum height.
We know that
v y ² = u y ² + 2a y s
Here u y = u sin 0, a y = – g, s = h max
v y = 0
Expression for horizontal range:
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range.
Horizontal range = Horizontal component of velocity x time of flight
R = u cos θ x t f → (1)
Time of flight (t f ) is the time taken by the projectile from point of projection to point the projectile hits the ground again
w.k.t = S y = u y t f + 1/2 a y t² f )
Here S y = 0 u y = u sin θ, a y = – g
0 = u sin θ t f – 1/2g t² f
1/2 gt t² f = u sin θ t f

In uniform circular motion the velocity vector turns continuously with out changing its magnitude. ie speed remains constant and direction changes. Even though the velocity is tangential to every point is a circle, the acceleration it acting towards the centre of the circle along the radius. This is called centripetal acceleration
Expression:
The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors. Let the directions of position and velocity vectors shift through same angle θ in a small time interval ∆t
For uniform circular motion r = \(\left|\bar{r}_{1}\right|\) = \(\left|\bar{r}_{2}\right|\)
and v = \(\left|\bar{v}_{1}\right|\) = \(\left|\bar{v}_{2}\right|\)
If the particle moves from position vector \(\bar{r}_{1}\) to \(\bar{r}_{2}\) the displacement is given by \(\overrightarrow{\Delta r}\) = \(\bar{r}_{2}\) – \(\bar{r}_{1}\)
and change in velocity from \(\bar{v}_{1}\) to \(\bar{v}_{2}\) is given ∆\(\bar { v }\) = \(\bar{v}_{2}\) – \(\bar{v}_{1}\)
The magnitudes of the displacement ∆r and ∆v satisfy the following relation
Here negative sign indicates that ∆v points radially inwards, towards the centre of the circle
For uniform circular motion v=rω where ω is the angular velocity of the particle about the center
The centripetal acceleration a = ω²r.

If the velocity changes both in speed and direction during circular motion, then we get non-uniform circular motion. Whenever the speed is not the same in a circular motion then the particle will have both centripetal and tangential acceleration.
The resultant acceleration is obtained by the vector sum of centripetal and tangential acceleration
Let the tangential acceleration be a t.
Centripetal acceleration is v²/r.
The magnitude of the resultant acceleration is a R = \(\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)
IV. Exercises:

(a) When the body starts from rest and moves with uniform acceleration is constant
(b) This graph represents, for a body moving with a uniform velocity or constant velocity. The zero slope of curve indicates zero acceleration.
(c) This v-t graph is a straight line not passing through origin indicates the body has a constant acceleration but greater than fig(i) as slope is more than the first one (more steeper)
(d) Greater changes in velocity (velocity variations are taking place in equal as travels of time. The graph indicates increasing acceleration.

From o to A(o to Is):
At t = os the particle has a zero velocity at t > 0 the particle has a negative velocity and moves in positive x-direction the slope dr/dt is negative. The particle is decelerating. Thus the velocity decreases during this time interval.
From A to B (Is to 2s):
From time Is to 2s the velocity increase and slope dv/dt becomes positive. The particle is accelerating. The velocity increases in this time interval.
From B to C (2s to 5s):
From 2s to 5s the velocity stays constant at 1 m/s. The acceleration is zero.
From C to D (6s to 7s):
From 5s to 6s the velocity decreases. Slope dv/dt is negative. The particle is decelerating. The velocity decreases to zero. The body comes to rest at 6s.
From D to E (6s to 7s)
The particle is at rest during this time interval.
Displacement: in 0 – 2s:
The total area under the curve from 0 to 2s displacement = 1/2bh + 1/2bh
=1/2 x 1.5 x (- 2) + 0.5 x 1
= – 1.5 + 0.25
= – 1.25 m
Distance: is 0 – 2s
The distance covered is = 1.5 + 0.25 = 1.75 m

:
Speed of water = V
(Range) max = radius = u 2 /g = v²/g
This range becomes the radius = (v²/g) of the circle where water sprinkled.
Area covered = Area of circle
= πr² = π\(\left(\frac{v^{2}}{g}\right)\)²
= π \(v^{4} / g^{2}\)

R = – (sin 2θ)
∵ the initial velocity and angle of projection are constants
R ∝ \(\frac { 1 }{ g }\)
g ∝ \(\frac { 1 }{ R }\)
According to acceleration, due to gravity In ascending order, the solution is. Mercury, Mars, Earth, Jupiter

We can resolve all vectors in x, y, z components w.r.t. Cartesian co-ordinate system. After resolving the components separately equate x components on both sides y components on both sides and z components on both side we get.
(a) T\(\hat{j}\) – mg\(\hat{j}\) = ma\(\hat{j}\)
T – mg = ma
(b) \(\overline{T}\) + \(\overline{F}\) = \(\overline{A}\) + \(\overline{B}\)
T x + F x = A x + B x
(c) \(\overline{T}\) – \(\overline{F}\) = \(\overline{A}\) – \(\overline{B}\)
T x – F x = A x – B x
(d) T\(\hat{j}\) + mg\(\hat{j}\) = ma\(\hat{j}\)
T + mg = ma

ω = θ/t θ = wt
In 24 hours, angular displacement made
θ = 360° (or) 2π rad
In 1 hours, angular displacement
θ = \(\frac { 360° }{ 24 }\)
θ = 15°
In radian θ = \(\frac { 2π }{ 24 }\) = \(\frac { π }{ 12 }\) radians.
θ = \(\frac { π }{ 12 }\) rad.

:
ω = π/12 rad/s
ω = θ/t
θ = w x t = π/12 x 4
θ = π/3 radian
θ = \(\frac { 180° }{ 3 }\)
= 60°

:
Angle described in 27 days = 2π rad = 360° days
Angie described in one day = 2π/27 radian
= \(\frac { 360° }{ 27 }\)
θ = 13.3°

∝ = 0.2 rad/s²
θ = ? t = 3s.
w 0 = 0
w.k.T θ = ω 0 t + 1/2 ∝ t²
θ = 0 + 1/2 x 0.2 x 9
θ = 0.9 rad
θ = 0 = 0.9 x 57.295° = 51°
The magnitude of the resultant vector R is given by
11th Physics Guide Kinematics Additional Important Questions and Answers
I. Multiple choice questions: