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CBSE NOTES CLASS 11 PHYSICS CHAPTER 3

MOTION IN A STRAIGHT LINE

Kinematics

Cartesian coordinate system

Reference frame

One dimensional motion or rectilinear motion

Two dimensional motion

Three dimensional motion

Path length

Displacement

Uniform Motion

Average speed

Uniform speed

Instantaneous speed

Velocity

Average velocity

Instantaneous velocity

Acceleration

v-t diagrams for uniformely accelerated motion

Positive acceleration

Negative acceleration

Kinematic Equations for Uniformly Accelerated Motion along a Straight Line

Proof of Kinematic Equations using Graphical method

Proof of Kinematic Equations using Analytical (Calculus) Method

Free fall

Distance travelled - Snth

Galileo’s law of odd numbers

Stopping distance of vehicles and reaction time

Relative Motion

Objects travelling in the same direction

Objects travelling in the opposite direction

Finding the relative velocity, when the particles are not moving collinearly

CBSE NOTES CLASS 11 PHYSICS CHAPTER 3

MOTION IN A STRAIGHT LINE

Kinematics

The branch of physics in which, we study ways to describe motion without going into the causes of motion.

Cartesian coordinate system

A coordinate system in which the coordinates of a point are its distances from a set of perpendicular lines that intersect at an origin, is called Cartesian coordinate system. The point of intersection of these three axes is called origin (O) and serves as the reference point.

Reference frame

The coordinate system (for measurement of positon) along with a clock (for measurement of time) constitutes a frame of reference.

One dimensional motion or rectilinear motion

Motion of objects along a straight line is called rectilinear motion. The object can move forward or backward, up or down etc.

The motion of an object is said to be one dimensional motion if only one out of three coordinates specifying the position of the object change with time.

Two dimensional motion

The motion of an object is said to be two dimensional if two out of three coordinates specifying the position of the object change with time. In such motion the object moves in a plane.

Three dimensional motion

The motion is said to be three dimensional motion if all the three coordinates specifying the position of an object change with respect to time, in such a motion an object moves in space.

Path length

The distance travelled by an object, without considering the direction, is called the path length. It is a scalar quantity.

Displacement

A displacement is the shortest distance from the initial to the final position of an object. It is a vector quantity. A displacement vector represents the length and direction of the straight line joining the initial and final point.

Let x1 and x2 be the positions of an object at time t1 and t2. The displacement, in time Δt = (t2 - t1), is given by the difference between the final and initial positions.

Δx = x2x1

Uniform Motion

If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line.

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Average speed

Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place

Average speed = Total path lengthTotal time interval

Uniform speed

If an object covers equal distances in equal intervals of time than the speed of the moving object is called uniform speed. In this type of motion, position – time graph is always a straight line.

Instantaneous speed

The speed of an object at any particular instant of time is called instantaneous speed. When a body is moving with uniform speed its

instantaneous speed = average speed = uniform speed.

Velocity: The rate of change of position or displacement of an object in a particular direction with respect to time is called velocity. It is equal to the displacement covered by an object per unit time.

It is a vector quantity.

It can be +ve, -ve or zero.

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Average velocity is defined as the change in position or displacement (Δx) divided by the time intervals (Δt), in which the displacement occurs,

 v = x2-x1t2- t1= ΔxΔt

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Instantaneous Velocity

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The velocity at an instant is defined as the limit of the average velocity as the time interval Δt becomes infinitesimally small.

It is the rate of change of position with respect to time, at that instant.

v = limΔt  0 ΔxΔt = dxdt= differential of x with respect to t

Acceleration

The rate of change of velocity of an object with respect to time is called its acceleration.

The average acceleration a over a time interval is defined as the change of velocity divided by the time interval

a = v-ut2- t1= ΔvΔt

It is a vector quantity

Its SI unit is meter/sec2 and its dimensions are [M0L1T-2].

It may be positive, negative or zero.

Positive acceleration

If the velocity of an object increases with time, its acceleration is positive, that is x and a have same direction.

Negative acceleration

If the velocity of an object decreases with time, its acceleration is negative. The negative acceleration is also called retardation or deceleration.

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v-t diagrams for uniformely accelerated motion

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Kinematic Equations for Uniformly Accelerated Motion along a Straight Line

  • If a = 0, then v = u, and s = ut
  1. v = u + at
  1. s = ut + ½ at2
  1. v2 – u2 = 2as

Proofs (Graphical method)

The area under the v - t curve represents the displacement over a given time interval.

The area under the a - t curve represents the change in velocity (v - u).

Consider motion of an object under uniform acceleration, as depicted in the diagram. From the diagram, we have,

OA = u, DB = v,

BC = v - u

OD = OA = t

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  1. We know that
    a = v - ut,  so, v=u+at

  2. Displacement s = Area between instants 0 and t

    = Area of rectangle OACD + Area of triangle ABC

    = OD × AO + ½ × AC × BC

    = u t + ½ × t × (v-u)

    Now since v - u = at (from first equation), we have

    s = u t + ½ × t × a t or

    s = u t + ½ a t2


  3. Displacement s = Area between instants 0 and t

    s = Area of trapezium OABD

    =OA+BD2× OC 

    s =v+u2×t 

    Now since t = (v-u)/a (from first equation), we have

    s =v+u2×t=v+uv-u2a

    s =v2- u22a             v2- u2=2as

Proof of Kinematic Equations using Analytical (Calculus) Method

We know that,

a=dvdt    dv=a dt

Integrating both sides

uvdv= 0ta dt

  = a0t dt since a is constant

    v-u=at       v=u+at

Also,

v=dxdt    dx=v dt   

   dx=u+at dt

Integrating both sides

x0xdx= 0tu+at dt

  = u0t dt+ a0tt dt

since u and a are constants

xx0 = ut + ½ at2

If x0 = 0 and we take x as s,

s = ut + ½ at2

Further

a =dvdt=dvdx.dxdt=vdvdx

a dx = v dv

Integrating both the sides,

xoxa dx= uvv dv

 ax-xo=v2- u22

 2 ax-xo=v2- u2 

Free fall

In the absence of the air resistance all bodies fall with the same acceleration towards earth from a small height. This is called free fall. The acceleration with which a body falls is called gravitational acceleration (g). Its value is 9.8 m/sec2. The equations for a free fall become,

v = - g t

s = - ½ g t2

v2 = 2 g h

Velocity after falling a vertical distance h,

v =2gh

Distance travelled

Motion under uniform accelertion

After (n - 1)th second sn-1 = u (n-1) + ½ a (n -1)2

After nth second sn = u n + ½ an2,

During nth second snth = sn - sn-1 = u + ½ a (2n -1)

Under free fall

After (n - 1)th second, sn-1 = ½ g (n -1)2

After nth second, sn = ½ gn2,

During nth second, snth = sn - sn-1 = ½ g (2n -1)

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Galileo’s law of odd numbers: “The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it.

  • Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have,

    s = ½ gt2

    After different time intervals, 0, τ, 2τ, 3τ… can be tabulated as follows,

    t

    s

    Distance in terms of

    so = ½ gτ 2

    Distance between successive intervals

    Ratio of distances traversed

    0

    0

    0

    τ

    ½ g × (τ)2 = ½ gτ 2

    so

    so

    1

    ½ g × (2τ)2 = 4 × ½ gτ 2

    4so

    3so

    3

    ½ g (3τ)2 = 9 × ½ gτ 2

    9so

    5so

    5

    ½ g (4τ)2 = 16 × ½ gτ 2

    16so

    7so

    7

    ½ g (5τ)2 = 25 × ½ gτ 2

    25so

    9so

    9

    ½ g (6τ)2 = 36 × ½ gτ 2

    36so

    11so

    11

Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping, is called stopping distance. We can find it by taking v = 0 and acceleration, a, to be negative. That is

s =u22a

  • The stopping distance is proportional to the square of the initial velocity.

Reaction time: Reaction time is the time a person takes to observe, think and act. The reaction time for catching an object,

tr=2dg

Relative Motion

The rate of change of distance of one object with respect to the other is called relative velocity. The relative velocity of an object B with respect to the object A when both are in motion is the rate of change of position of object B with respect to the object A.

Relative velocity of object A with respect to object B,

vAB= vA+ -vB= vA -vB

and vBA= vB+ -vA= vB -vA

When both objects are moving in the same direction, the relative velocity of object A with respect to the object B,

vAB = vA – vB

When the objects are moving in opposite direction,

vAB = vA + vB

  • If vA=vB,

    vA -vB = 0,

    xA(t) -xB(t) = xA(0) -xB(0).

    Therefore, the two objects stay at a constant distance, and their position–time graphs are straight lines parallel to each other.

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  • If vA<vB, and have same sign

    vA -vB < 0 (negative),

    Therefore, one graph is steeper than the other and they meet at a common point.

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  • If vA and vB are of opposite signs and the objects move towards each other, one of the graphs has +ve and another –ve slope and they meet at a common point again. On the other hand if the objects are moving in opposite direction, they will never meet.

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Finding the relative velocity, when the particles are not moving collinearly.

While finding the relative velocity, velocity vectors of two objects need to be added as follows.

Relative velocity of body A with respect to body B is given by,

vAB= vA+ -vB

Magnitude of the relative velocity is given by,

vAB= vA2+ vB2- 2 vAvBcos

Where θ is the angle between the directions of motion of A and B.

And tan β= vBsinvA-vBcos

Where β is the angle of resulting velocity vector from the direction of vA

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