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 x_{1} and x_{2} be the positions of an object at time t_{1} and t_{2}. The displacement, in time Δt = (t_{2 } t_{1}), is given by the difference between the final and initial positions.
Δx = x_{2} – x_{1}
 If x_{2} > x_{1}, Δx is positive; and if x_{2} < x_{1}, Δx is negative.
 The magnitude of displacement is less than or equal to the actual distance (path length) travelled by the object in the given time interval.
Displacement ≤ Path length
The two quantities are equal only if the object does not change its direction during the course of its motion.
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.
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
$$\mathrm{Average\; speed\; =}\frac{\mathrm{Total\; path\; length}}{\mathrm{Total\; time\; interval}}$$
 Speed is a scalar quantity.
 Its SI unit is meter/sec. Dimensional formula of speed is [M^{0}L^{1} T^{1}].
 It is never negative.
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.
Average velocity is defined as the change in position or displacement (Δx) divided by the time intervals (Δt), in which the displacement occurs,
$$\mathrm{}\stackrel{\u203e}{\mathrm{v}}\mathrm{}=\mathrm{}\frac{{\mathrm{x}}_{2}{\mathrm{x}}_{1}}{{\mathrm{t}}_{2}\mathrm{}{\mathrm{t}}_{1}}=\mathrm{}\frac{\mathrm{\Delta}\mathrm{x}}{\mathrm{\Delta}\mathrm{t}}$$
 Geometrically, the average velocity is the slope of the straight line P_{1}P_{2} connecting the initial position P_{1} to the final position P_{2}
 The SI unit for velocity is ms^{–1}. It is a vector quantity. Its dimensional formula is [M^{0}L^{1}T^{1}].
 It may be negative, positive or zero.
 When a body moves in a straight line and in the same direction, the average speed and is equal to the magnitude of average velocity.
Instantaneous Velocity
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.
$$\mathrm{v}\mathrm{}=\mathrm{}\underset{\mathit{\Delta t}\mathrm{}\mathrm{}0\mathrm{}}{\mathit{lim}}\frac{\mathrm{\Delta}\mathrm{x}}{\mathrm{\Delta}\mathrm{t}}\mathrm{}=\mathrm{}\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}=\mathrm{}\mathrm{differential\; of\; x\; with\; respect\; to\; t}$$
 For uniform motion, velocity is the same as the average velocity at all instants.
 Instantaneous speed or simply speed is the magnitude of instantaneous velocity or simply velocity.
Acceleration
The rate of change of velocity of an object with respect to time is called its acceleration.
The average acceleration $\stackrel{\u203e}{\mathit{a}}$ over a time interval is defined as the change of velocity divided by the time interval
$$\stackrel{\u203e}{\mathit{a}}\mathrm{}=\mathrm{}\frac{\mathrm{v}\mathrm{u}}{{\mathrm{t}}_{2}\mathrm{}{\mathrm{t}}_{1}}=\mathrm{}\frac{\mathrm{\Delta}\mathrm{v}}{\mathrm{\Delta}\mathrm{t}}$$
It is a vector quantity
Its SI unit is meter/sec^{2 }and its dimensions are [M^{0}L^{1}T^{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.
vt diagrams for uniformely accelerated motion
Kinematic Equations for Uniformly Accelerated Motion along a Straight Line
 If a = 0, then v = u, and s = ut



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
 We know that
$$\mathrm{a}\mathrm{}=\mathrm{}\frac{\mathrm{v}\mathrm{}\mathrm{}\mathrm{u}}{\mathrm{t}},\mathrm{}\mathrm{}\mathrm{s}\mathrm{o},\mathrm{}\mathrm{v}=\mathrm{u}+\mathrm{a}\mathrm{t}$$  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 × (vu)
Now since v  u = at (from first equation), we have
s = u t + ½ × t × a t or
⇒ s = u t + ½ a t^{2}
 Displacement s = Area between instants 0 and t
$$\Rightarrow \mathrm{s\; =\; Area\; of\; trapezium\; OABD}$$
$$=\frac{\mathrm{O}\mathrm{A}+\mathrm{B}\mathrm{D}}{2}\times \mathrm{}\mathrm{O}\mathrm{C}\mathrm{}$$
$$\Rightarrow \mathrm{s}\mathrm{}=\frac{\mathrm{v}+\mathrm{u}}{2}\times \mathrm{t}\mathrm{}$$
Now since t = (vu)/a (from first equation), we have
$$\mathrm{s}\mathrm{}=\frac{\mathrm{v}+\mathrm{u}}{2}\times \mathrm{t}=\frac{\left(\mathrm{v}+\mathrm{u}\right)\left(\mathrm{v}\mathrm{u}\right)}{2\mathrm{a}}$$
$$\mathrm{s}\mathrm{}=\frac{{\mathrm{v}}^{2}\mathrm{}{\mathrm{u}}^{2}}{2\mathrm{a}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\mathrm{v}}^{2}\mathrm{}{\mathrm{u}}^{2}=2\mathrm{a}\mathrm{s}$$
Proof of Kinematic Equations using Analytical (Calculus) Method
We know that,
$$\mathrm{a}=\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{t}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{d}\mathrm{v}=\mathrm{a}\mathrm{}\mathrm{d}\mathrm{t}$$
Integrating both sides
$${\int}_{\mathrm{u}}^{\mathrm{v}}\mathrm{d}\mathrm{v}=\mathrm{}{\int}_{0}^{\mathrm{t}}\mathrm{a}\mathrm{}\mathrm{d}\mathrm{t}$$
$$\mathrm{}\mathrm{}=\mathrm{}\mathrm{a}{\int}_{0}^{\mathrm{t}}\mathrm{}\mathrm{d}\mathrm{t}\mathrm{}\left(\mathrm{since\; a\; is\; constant}\right)$$
$$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{v}\mathrm{u}=\mathrm{a}\mathrm{t}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{v}=\mathrm{u}+\mathrm{a}\mathrm{t}$$
Also,
$$\mathrm{v}=\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{d}\mathrm{x}=\mathrm{v}\mathrm{}\mathrm{d}\mathrm{t}\mathrm{}\mathrm{}\mathrm{}$$
$$\mathrm{}\mathrm{}\mathrm{}\mathrm{d}\mathrm{x}=\left(\mathrm{u}+\mathrm{a}\mathrm{t}\right)\mathrm{}\mathrm{d}\mathrm{t}$$
Integrating both sides
$${\int}_{\mathrm{x}0}^{\mathrm{x}}\mathrm{d}\mathrm{x}=\mathrm{}{\int}_{0}^{\mathrm{t}}\left(\mathrm{u}+\mathrm{a}\mathrm{t}\right)\mathrm{}\mathrm{d}\mathrm{t}$$
$$\mathrm{}\mathrm{}=\mathrm{}\mathrm{u}{\int}_{0}^{\mathrm{t}}\mathrm{}\mathrm{d}\mathrm{t}+\mathrm{}\mathrm{a}{\int}_{0}^{\mathrm{t}}\mathrm{t}\mathrm{}\mathrm{d}\mathrm{t}$$
$$\left(\mathrm{since\; u\; and\; a\; are\; constants}\right)$$
x – x_{0} = ut + ½ at^{2}
If x_{0} = 0 and we take x as s,
s = ut + ½ at^{2}
Further
$$\mathrm{a}\mathrm{}=\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}.\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}=\mathrm{v}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}$$
⇒ a dx = v dv
Integrating both the sides,
$${\int}_{{\mathrm{x}}_{\mathrm{o}}}^{\mathrm{x}}\mathrm{a}\mathrm{}\mathrm{d}\mathrm{x}=\mathrm{}{\int}_{\mathrm{u}}^{\mathrm{v}}\mathrm{v}\mathrm{}\mathrm{d}\mathrm{v}$$
$$\mathrm{\Rightarrow}\mathrm{}\mathrm{a}\left(\mathrm{x}{\mathrm{x}}_{\mathrm{o}}\right)=\frac{{\mathrm{v}}^{2}\mathrm{}{\mathrm{u}}^{2}}{2}$$
$$\mathrm{}2\mathrm{}\mathrm{a}\left(\mathrm{x}{\mathrm{x}}_{\mathrm{o}}\right)={\mathrm{v}}^{2}\mathrm{}{\mathrm{u}}^{2}\mathrm{}$$
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/sec^{2}. The equations for a free fall become,
v =  g t
s =  ½ g t^{2}
v^{2} = 2 g h
Velocity after falling a vertical distance h,
$\mathrm{v}\mathrm{}=\sqrt{2\mathrm{g}\mathrm{h}}$
Distance travelled
Motion under uniform accelertion
After (n  1)th second s_{n1} = u (n1) + ½ a (n 1)^{2}
After nth second s_{n} = u n + ½ an^{2},
During nth second s_{nth} = s_{n}  s_{n1} = u + ½ a (2n 1)
Under free fall
After (n  1)th second, s_{n1} = ½ g (n 1)^{2}
After nth second, s_{n} = ½ gn^{2},
During nth second, s_{nth} = s_{n}  s_{n1} = ½ g (2n 1)
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 gt^{2}
After different time intervals, 0, τ, 2τ, 3τ… can be tabulated as follows,
t
s
Distance in terms of
s_{o} = ½ gτ^{ 2}
Distance between successive intervals
Ratio of distances traversed
0
0
0
τ
½ g × (τ)^{2} = ½ gτ^{ 2}
s_{o}
s_{o}
1
2τ
½ g × (2τ)^{2} = 4 × ½ gτ^{ 2}
4s_{o}
3s_{o}
3
3τ
½ g (3τ)^{2} = 9 × ½ gτ^{ 2}
9s_{o}
5s_{o}
5
4τ
½ g (4τ)^{2 }= 16 × ½ gτ^{ 2}
16s_{o}
7s_{o}
7
5τ
½ g (5τ)^{2} = 25 × ½ gτ^{ 2}
25s_{o}
9s_{o}
9
6τ
½ g (6τ)^{2 }= 36 × ½ gτ^{ 2}
36s_{o}
11s_{o}
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
$$\mathrm{s}\mathrm{}=\frac{{\mathrm{u}}^{2}}{2\mathrm{a}}$$
 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,
$${\mathrm{t}}_{\mathrm{r}}=\sqrt{\frac{2\mathrm{d}}{\mathrm{g}}}$$
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,
$\overrightarrow{{\mathrm{v}}_{\mathrm{A}\mathrm{B}}}=\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}+\mathrm{}\left(\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}\right)=\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}$
and $\overrightarrow{{\mathrm{v}}_{\mathrm{B}\mathrm{A}}}=\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}+\mathrm{}\left(\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}\right)=\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}$
When both objects are moving in the same direction, the relative velocity of object A with respect to the object B,
v_{AB} = v_{A} – v_{B}
When the objects are moving in opposite direction,
v_{AB} = v_{A} + v_{B}
 If $\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}=\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}$,
$\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}$ = 0,
$\overrightarrow{{\mathrm{x}}_{\mathrm{A}}}\left(\mathrm{t}\right)\mathrm{}\overrightarrow{{\mathrm{x}}_{\mathrm{B}}}\left(\mathrm{t}\right)$ = $\overrightarrow{{\mathrm{x}}_{\mathrm{A}}}\left(0\right)\mathrm{}\overrightarrow{{\mathrm{x}}_{\mathrm{B}}}\left(0\right)$.
Therefore, the two objects stay at a constant distance, and their position–time graphs are straight lines parallel to each other.
 If $\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}<\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}$, and have same sign
$\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}$ < 0 (negative),
Therefore, one graph is steeper than the other and they meet at a common point.
 If $\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}$ 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.
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,
$$\overrightarrow{{\mathrm{v}}_{\mathrm{A}\mathrm{B}}}=\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}+\mathrm{}\left(\overrightarrow{{\mathrm{v}}_{\mathrm{B}}}\right)$$
Magnitude of the relative velocity is given by,
$${\mathrm{v}}_{\mathrm{A}\mathrm{B}}=\mathrm{}\sqrt{{{\mathrm{v}}_{\mathrm{A}}}^{2}+\mathrm{}{{\mathrm{v}}_{\mathrm{B}}}^{2}\mathrm{}2\mathrm{}{\mathrm{v}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{B}}\mathrm{cos}}$$
Where θ is the angle between the directions of motion of A and B.
$$\mathrm{A}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{}\mathrm{\beta}=\mathrm{}\frac{{\mathrm{v}}_{\mathrm{B}}\mathrm{sin}}{{\mathrm{v}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{B}}\mathrm{cos}}$$
Where β is the angle of resulting velocity vector from the direction of $\overrightarrow{{\mathrm{v}}_{\mathrm{A}}}$