Classification and Nomenclature of Organic Compounds
Classification of organic compounds
Organic compounds have been classified on the basis of carbon skeleton (structure) or functional groups or the concept of homology.
(1) Classification based on structure
(i) Acyclic or open-chain compounds: Organic compounds in which all the carbon atoms are linked to one another to form open chains (straight or branched) are called acyclic or open chain compounds. These may be either saturated or unsaturated. For example,
\[\underset{\text{Butane}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\]
\[\underset{\text{Isobutane}}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C{{H}_{3}}-C-C\equiv CH}}\,}}\,}}\,\]
These compounds are also called as aliphatic compounds.
(ii) Cyclic or closed-chain compounds: Cyclic compounds contain at least one ring or closed chain of atoms. The compounds with only one ring of atoms in the molecule are known as monocyclic but those with more than one ring of atoms are termed as polycyclic. These are further divided into two subgroups.
(a) Homocyclic or carbocyclic: These are the compounds having a ring or rings of carbon atoms only in the molecule. The carbocyclic or homocyclic compounds may again be divided into two types:
Alicyclic compounds: These are the compounds which contain rings of three or more carbon atoms. These resemble with aliphatic compounds than aromatic compounds in many respects. That is why these are named alicyclic, i.e., aliphatic cyclic. These are also termed as polymethylenes. Some of the examples are,
Aromatic compounds: These compounds consist of at least one benzene ring, i.e., a six-membered carbocyclic ring having alternate single and double bonds. Generally, these compounds have some fragrant odour and hence, named as aromatic (Greek word aroma meaning sweet smell).
These are also called benzenoid aromatics.
Note:q Non-benzenoid aromatics: There are aromatic compounds, which have structural units different from benzenoid type and are known as Non-benzenoid aromatics e.g. Tropolone, azulene etc.
(b) Heterocyclic compounds: Cyclic compounds containing one or more hetero atoms (e.g. O, N, S etc.) in the ring are called heterocyclic compounds. These are of two types:
Alicyclic heterocyclic compounds: Heterocyclic compounds which resemble aliphatic compounds in their properties are called Alicyclic heterocyclic compounds. For example,
Aromatic heterocyclic compounds: Heterocyclic compounds which resemble benzene and other aromatic compounds in most of their properties are called Aromatic heterocyclic compounds. For example,
(2) Classification based on functional groups: A functional group is an atom or group of atoms in a molecule that gives the molecule its characteristic chemical properties. more...
Distance and Displacement
1) Distance: It is the actual path length covered by a moving particle in a given interval of time.
(i) If a particle starts from A and reach to C through point B as shown in the figure.
Then distance travelled by particle \[=AB+BC=7\,m\]
(ii) Distance is a scalar quantity.
(iii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]\]
(iv) Unit: metre (S.I.)
(2) Displacement: Displacement is the change in position vector i.e., A vector joining initial to final position.
(i) Displacement is a vector quantity
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]\]
(iii) Unit: metre (S.I.)
(iv) In the above figure the displacement of the particle \[\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\]
\[\Rightarrow \,\,\left| AC \right|=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}+2(AB)\,(BC)\,\cos \,{{90}^{o}}\,}=\,\,5\,m\]
(v) If \[{{\vec{S}}_{1}},\,{{\vec{S}}_{2}},\,{{\vec{S}}_{3}}\,........\,{{\vec{S}}_{n}}\] are the displacements of a body then the total (net) displacement is the vector sum of the individuals. \[\vec{S}={{\vec{S}}_{1}}+\,{{\vec{S}}_{2}}+\,{{\vec{S}}_{3}}+\,........\,+{{\vec{S}}_{n}}\]
(3) Comparison between distance and displacement:
(i) The magnitude of displacement is equal to minimum possible distance between two positions.
So \[distance\ge \left| Displacement \right|\].
(ii) For a moving particle distance can never be negative or zero while displacement can be.
(zero displacement means that body after motion has came back to initial position)
i.e., Distance > 0 but Displacement > = or < 0
(iii) For motion between two points displacement is single valued while distance depends on actual path and so can have many values.
(iv) For a moving particle distance can never decrease with time while displacement can. Decrease in displacement with time means body is moving towards the initial position.
(v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without change in direction.
(vi) If \[{{\vec{r}}_{A}}\,\,and\,\,{{\vec{r}}_{B}}\] are the position vectors of particle initially and finally.
Then displacement of the particle
\[{{\vec{r}}_{AB}}={{\vec{r}}_{B}}-{{\vec{r}}_{A}}\]
and s is the distance travelled if the particle has gone through the path APB.
Sample problems based on distance and displacement
Problem 1. A man goes 10 m towards North, then 20m towards east then displacement is
[KCET (Med.) 1999; JIPMER 1999; AFMC 2003]
(a) 22.5 m (b) 25 m (c) 25.5 m (d) 30 m
Solution: (a) If we take east as \[x-\] axis and north as \[y-\] axis, then displacement \[=20\,\hat{i}\,+\,10\hat{j}\]
So, magnitude of displacement \[=\sqrt{{{20}^{2}}+{{10}^{2}}}=10\sqrt{5}=\text{ }22.5\,\,m.\]
Problem 2. A body moves over one fourth of a circular arc in a circle of radius r. The magnitude of distance travelled and displacement will be respectively
(a) \[\frac{\pi r}{2},\,r\sqrt{2}\] (b) \[\frac{\pi r}{4},\,r\] (c) \[\pi r,\,\frac{r}{\sqrt{2}}\] (d) \[\pi r,\,r\]
Solution: (a) Let particle start from A, its position vector \[{{\vec{r}}_{O}}_{A}=r\hat{i}\]
After one quarter position vector \[{{\vec{r}}_{OB}}=r\,\hat{j}.\]
So displacement \[=r\hat{j}-r\hat{i}\]
Magnitude of displacement \[=r\sqrt{2}.\]
and distance = one fourth of circumference \[=\frac{2\pi r}{4}=\frac{\pi r}{2}\]
Problem 3. The displacement of the point of the wheel initially in contact with the ground, when more...
Rest and Motion
If a body does not change its position as time passes with respect to frame of reference, it is said to be at rest. And if a body changes its position as time passes with respect to frame of reference, it is said to be in motion. Frame of Reference: It is a system to which a set of coordinates are attached and with reference to which observer describes any event. A passenger standing on platform observes that tree on a platform is at rest. But when the same passenger is passing away in a train through station, observes that tree is in motion. In both conditions observer is right. But observations are different because in first situation observer stands on a platform, which is reference frame at rest and in second situation observer moving in train, which is reference frame in motion. So rest and motion are relative terms. It depends upon the frame of references.
Speed and Velocity
(1) Speed: Rate of distance covered with time is called speed.
(i) It is a scalar quantity having symbol \[\upsilon \].
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{1}} \right]\]
(iii) Unit: metre/second (S.I.), cm/second (C.G.S.)
(iv) Types of speed:
(a) Uniform speed: When a particle covers equal distances in equal intervals of time, (no matter how small the intervals are) then it is said to be moving with uniform speed. In given illustration motorcyclist travels equal distance (= 5 m) in each second. So we can say that particle is moving with uniform speed of 5 m/s.
(b) Non-uniform (variable) speed: In non-uniform speed particle covers unequal distances in equal intervals of time. In the given illustration motorcyclist travels 5m in 1st second, 8m in 2nd second, 10m in 3rd second, 4m in 4th second etc.
Therefore its speed is different for every time interval of one second. This means particle is moving with variable speed.
(c) Average speed: The average speed of a particle for a given ‘Interval of time’ is defined as the ratio of distance travelled to the time taken.
\[Average\text{ }speed=\frac{\text{Distance travelled}}{\text{Time taken}}\]; \[{{v}_{av}}=\frac{\Delta s}{\Delta t}\]
q Time average speed: When particle moves with different uniform speed \[{{\upsilon }_{1}},\,\,{{\upsilon }_{2}},\,\,{{\upsilon }_{3}}\]... etc. in different time intervals\[{{t}_{1}},\,\,{{t}_{2}},\,\,{{t}_{3}}\], ... etc. respectively, its average speed over the total time of journey is given as
\[{{v}_{av}}=\frac{\text{Total distance covered}}{\text{Total time elapsed}}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}=\frac{{{\upsilon }_{1}}{{t}_{1}}+{{\upsilon }_{2}}{{t}_{2}}+{{\upsilon }_{3}}{{t}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}\]
Special case: When particle moves with speed \[{{v}_{1}}\] upto half time of its total motion and in rest time it is moving with speed \[{{v}_{2}}\] then \[{{v}_{av}}=\frac{{{v}_{1}}+{{v}_{2}}}{2}\]
q Distance averaged speed: When a particle describes different distances \[{{d}_{1}},\,\,{{d}_{2}},\,\,{{d}_{3}}\], ...... with different time intervals \[{{t}_{1}},\,\,{{t}_{2}},\,\,{{t}_{3}}\], ...... with speeds \[{{v}_{1}},{{v}_{2}},{{v}_{3}}......\] respectively then the speed of particle averaged over the total distance can be given as
\[{{\upsilon }_{av}}=\frac{\text{Total distance covered}}{\text{Total time elapsed}}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{\frac{{{d}_{1}}}{{{\upsilon }_{1}}}+\frac{{{d}_{2}}}{{{\upsilon }_{2}}}+\frac{{{d}_{3}}}{{{\upsilon }_{3}}}+......}\]
q When particle moves the first half of a distance at a speed of v1 and second half of the distance at speed v2 then
\[{{v}_{av}}=\frac{2{{v}_{1}}{{v}_{2}}}{{{v}_{1}}+{{v}_{2}}}\]
q When particle covers one-third distance at speed v1, next one third at speed v2 and last one third at speed v3, then
\[{{v}_{av}}=\frac{3\,{{v}_{1}}{{v}_{2}}{{v}_{3}}}{{{v}_{1}}{{v}_{2}}+{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{1}}}\]
(d) Instantaneous speed: It is the speed of a particle at particular instant. When we say “speed”, it usually means instantaneous speed.
The instantaneous speed is average speed for infinitesimally small time interval (i.e., \[\Delta t\to 0\]). Thus
Instantaneous speed \[v=\underset{\Delta t\to 0}{\mathop{\lim }}\,\,\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt}\]
(2) Velocity: Rate of change of position i.e. rate of displacement with time is called velocity.
(i) It is a scalar quantity having symbol \[v\].
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{1}} \right]\]
(iii) Unit: metre/second (S.I.), cm/second (C.G.S.)
(iv) Types
(a) Uniform velocity: A particle is said to have uniform velocity, if magnitudes as well as more...
Acceleration
The time rate of change of velocity of an object is called acceleration of the object.
(1) It is a vector quantity. It’s direction is same as that of change in velocity (Not of the velocity)
(2) There are three possible ways by which change in velocity may occur
When only direction of velocity changes
When only magnitude of velocity changes
When both magnitude and direction of velocity changes
Acceleration perpendicular to velocity
Acceleration parallel or anti-parallel to velocity
Acceleration has two components one is perpendicular to velocity and another parallel or anti-parallel to velocity
e.g. Uniform circular motion
e.g. Motion under gravity
e.g. Projectile motion
(3) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{2}} \right]\]
(4) Unit: metre/second2 (S.I.); cm/second2 (C.G.S.)
(5) Types of acceleration:
(i) Uniform acceleration: A body is said to have uniform acceleration if magnitude and direction of the acceleration remains constant during particle motion.
Note: q If a particle is moving with uniform acceleration, this does not necessarily imply that particle is moving in straight line. e.g. Projectile motion.
(ii) Non-uniform acceleration: A body is said to have non-uniform acceleration, if magnitude or direction or both, change during motion.
(iii) Average acceleration: \[{{\vec{a}}_{a\upsilon }}=\frac{\Delta \vec{v}}{\Delta \vec{t}}=\frac{{{{\vec{v}}}_{2}}-{{{\vec{v}}}_{1}}}{\Delta t}\]
The direction of average acceleration vector is the direction of the change in velocity vector as \[\vec{a}=\frac{\Delta \vec{v}}{\Delta t}\]
(iv) Instantaneous acceleration \[=\,\,\vec{a}=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta \vec{v}}{\Delta t}=\frac{d\vec{v}}{dt}\]
(v) For a moving body there is no relation between the direction of instantaneous velocity and direction of acceleration.
e.g. (a) In uniform circular motion \[\theta =90{}^\text{o}\] always
(b) In a projectile motion \[\theta \] is variable for every point of trajectory.
(vi) If a force \[\overrightarrow{F}\] acts on a particle of mass m, by Newton’s 2nd law, acceleration \[\vec{a}=\frac{{\vec{F}}}{m}\]
(vii) By definition \[\vec{a}=\frac{d\vec{v}}{dt}=\frac{{{d}^{2}}\vec{x}}{d{{t}^{2}}}\left[ \text{As}\,\,\vec{v}=\frac{d\vec{x}}{dt} \right]\]
i.e., if x is given as a function of time, second time derivative of displacement gives acceleration
(viii) If velocity is given as a function of position, then by chain rule \[a=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=v.\frac{d\upsilon }{dx}\,\left[ \text{as}\,\,v=\frac{dx}{dt} \right]\]
(ix) If a particle is accelerated for a time t1 by acceleration a1 and for time t2 by acceleration a2 then average acceleration is \[{{a}_{a\upsilon }}=\frac{{{a}_{1}}{{t}_{1}}+{{a}_{2}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\]
(x) If same force is applied on two bodies of different masses \[{{m}_{1}}\] and \[{{m}_{2}}\] separately then it produces accelerations \[{{a}_{1}}\,\,and\,\,{{a}_{2}}\] respectively. Now these bodies are attached together and form a combined system and same force is applied on that system so that a be the acceleration of the combined system, then
more...
Position Time Graph
During motion of the particle its parameters of kinematical analysis (u, v, a, r) changes with time. This can be represented on the graph.
Position time graph is plotted by taking time t along x-axis and position of the particle on y-axis.
Let AB is a position-time graph for any moving particle
As Velocity \[=\frac{\text{Change in position}}{\text{Time taken}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}}\] … (i)
From triangle ABC \[\tan \theta =\frac{BC}{AC}=\frac{AD}{AC}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}}\] …. (ii)
By comparing (i) and (ii) \[\operatorname{Velocity} = tan\,\theta \]
\[v = tan\,\theta \]
It is clear that slope of position-time graph represents the velocity of the particle.
Various position – time graphs and their interpretation
q = 0o so v = 0 i.e., line parallel to time axis represents that the particle is at rest.
q = 90o so v = ¥ i.e., line perpendicular to time axis represents that particle is changing its position but time does not changes it means the particle possesses infinite velocity. Practically this is not possible.
q = constant so v = constant, a = 0 i.e., line with constant slope represents uniform velocity of the particle.
\[\theta \] is increasing so v is increasing, a is positive. i.e., line bending towards position axis represents increasing velocity of particle. It means the particle possesses acceleration.
Velocity Time Graph
The graph is plotted by taking time t along x-axis and velocity of the particle on y-axis.
Distance and displacement: The area covered between the velocity time graph and time axis gives the displacement and distance travelled by the body for a given time interval.
Then Total distance \[=\,|{{A}_{1}}|\,+\,|{{A}_{2}}|\,+\,|{{A}_{3}}|\,\]
= Addition of modulus of different area. i.e. \[s\,=\,\int{|\upsilon |\,dt}\]
Total displacement \[={{A}_{1}}+{{A}_{2}}+{{A}_{3}}\]
= Addition of different area considering their sign. i.e. \[r\,=\,\int{\upsilon \,dt}\]
here \[{{\operatorname{A}}_{1}}\,\,and {{A}_{2}}\] are area of triangle 1 and 2 respectively and \[{{A}_{3}}\] is the area of trapezium.
Acceleration: Let AB is a velocity-time graph for any moving particle
As \[Acceleration\text{ }=\frac{\text{Change in velocity}}{\text{Time taken}}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}\] .… (i)
From triangle ABC, \[\tan \theta =\frac{BC}{AC}=\frac{AD}{AC}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}\] …. (ii)
By comparing (i) and (ii)
Acceleration (a) \[=\tan \theta \]
It is clear that slope of velocity-time graph represents the acceleration of the particle.
Various velocity – time graphs and their interpretation
q = 0, a = 0, v = constant i.e., line parallel to time axis represents that the particle is moving with constant velocity.
q = 90o, a = ¥, v = increasing i.e., line perpendicular to time axis represents that the particle is increasing its velocity, but time does not change. It means the particle possesses infinite acceleration. Practically it is not possible.
q =constant, so a = constant and v is increasing uniformly with time i.e., line with constant slope represents uniform acceleration of the particle.
Motion Under Gravity
The force of attraction of earth on bodies, is called force of gravity. Acceleration produced in the body by the force of gravity, is called acceleration due to gravity. It is represented by the symbol g.
In the absence of air resistance, it is found that all bodies (irrespective of the size, weight or composition) fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small altitude \[\left( h << R \right)\] is called free fall.
An ideal one-dimensional motion under gravity in which air resistance and the small changes in acceleration with height are neglected.
(1) If a body dropped from some height (initial velocity zero)
(i) Equation of motion: Taking initial position as origin and direction of motion (i.e., downward direction) as a positive, here we have
u = 0 [As body starts from rest]
a = +g [As acceleration is in the direction of motion]
v = g t … (i)
\[h=\frac{1}{2}g{{t}^{2}}\] … (ii)
\[{{\upsilon }^{2}}=2gh\] … (iii)
\[{{h}_{n}}=\frac{g}{2}(2n-1)\] ... (iv)
(ii) Graph of distance velocity and acceleration with respect to time: time:
(iii) As \[\operatorname{h} = \left( 1/2 \right)\,g{{t}^{2}}, i.e., h\propto {{t}^{2}}\], distance covered in time t, 2t, 3t, etc., will be in the ratio of \[{{1}^{2}}:\text{ }{{2}^{2}}:\text{ }{{3}^{2}}\], i.e., square of integers.
(iv) The distance covered in the nth sec, \[{{h}_{n}}=\frac{1}{2}g\,(2n-1)\]
So distance covered in I, II, III sec, etc., will be in the ratio of \[1 : 3 : 5\], i.e., odd integers only.
(2) If a body is projected vertically downward with some initial velocity
Equation of motion: \[\upsilon =u+g\,t\]
\[h=ut+\frac{1}{2}g\,{{t}^{2}}\]
\[{{\upsilon }^{2}}={{u}^{2}}+2gh\]
\[{{h}_{n}}=u+\frac{g}{2}\,(2n-1)\]
(3) If a body is projected vertically upward
(i) Equation of motion: Taking initial position as origin and direction of motion (i.e., vertically up) as positive
\[\operatorname{a} = g\] [As acceleration is downwards while motion upwards]
So, if the body is projected with velocity u and after time t it reaches up to height h then
\[\upsilon =u-g\,t\]; \[h=ut-\frac{1}{2}g\,{{t}^{2}}\]; \[{{\upsilon }^{2}}={{u}^{2}}-2gh;\,\,{{h}_{n}}=u-\frac{g}{2}\,(2n-1)\]
(ii) For maximum height \[\operatorname{v} = 0\]
So from above equation
\[\operatorname{u} = gt,\]
\[h=\frac{1}{2}g{{t}^{2}}\]
and \[{{u}^{2}}=2gh\]
(iii) Graph of distance, velocity and acceleration with respect to time (for maximum height):
It is clear that both quantities do not depend upon the mass of the body or we can say that in absence of air resistance, all bodies fall on the surface of the earth with the same rate.
(4) In case of motion under gravity for a given body, mass, acceleration, and mechanical energy remain constant while speed, more...
Equations of Kinematics
These are the various relations between u, v, a, t and s for the moving particle where the notations are used as:
u = Initial velocity of the particle at time t = 0 sec
v = Final velocity at time t sec
a = Acceleration of the particle
s = Distance travelled in time t sec
sn = Distance travelled by the body in nth sec
(1) When particle moves with zero acceleration
(i) It is a unidirectional motion with constant speed.
(ii) Magnitude of displacement is always equal to the distance travelled.
(iii) v = u, s = u t [As a = 0]
(2) When particle moves with constant acceleration
(i) Acceleration is said to be constant when both the magnitude and direction of acceleration remain constant.
(ii) There will be one dimensional motion if initial velocity and acceleration are parallel or anti-parallel to each other.
(iii) Equations of motion in scalar from Equation of motion in vector from
\[\upsilon =u+at\] \[\vec{v}=\vec{u}+\vec{a}t\]
\[s=ut+\frac{1}{2}a{{t}^{2}}\] \[\vec{s}=\vec{u}t+\frac{1}{2}\vec{a}{{t}^{2}}\]
\[{{\upsilon }^{2}}={{u}^{2}}+2as\] \[\vec{v}.\vec{v}-\vec{u}.\vec{u}=2\vec{a}.\vec{s}\]
\[s=\left( \frac{u+v}{2} \right)\,t\] \[\vec{s}=\frac{1}{2}(\vec{u}+\vec{v})\,t\]
\[{{s}_{n}}=u+\frac{a}{2}\,(2n-1)\] \[{{\vec{s}}_{n}}=\vec{u}+\frac{{\vec{a}}}{2}\,(2n-1)\]
(3) Important points for uniformly accelerated motion
(i) If a body starts from rest and moves with uniform acceleration then distance covered by the body in t sec is proportional to t2 (i.e. \[s\propto {{t}^{2}}\]).
So we can say that the ratio of distance covered in 1 sec, 2 sec and 3 sec is \[{{1}^{2}}:{{2}^{2}}:{{3}^{2}}\] or 1 : 4 : 9.
(ii) If a body starts from rest and moves with uniform acceleration then distance covered by the body in nth sec is proportional to \[(2n-1)\] (i.e. \[{{s}_{n}}\propto \,(2n-1)\]
So we can say that the ratio of distance covered in I sec, II sec and III sec is 1 : 3 : 5.
(iii) A body moving with a velocity u is stopped by application of brakes after covering a distance s. If the same body moves with velocity nu and same braking force is applied on it then it will come to rest after covering a distance of \[{{\operatorname{n}}^{2}}s\].
As \[{{\upsilon }^{2}}={{u}^{2}}-2as\Rightarrow 0={{u}^{2}}-2as\Rightarrow s=\frac{{{u}^{2}}}{2a},\,\,\,s\propto {{u}^{2}}\] [since a is constant]
So we can say that if u becomes n times then s becomes \[{{n}^{2}}\] times that of previous value.
(iv) A particle moving with uniform acceleration from A to B along a straight line has velocities \[{{\upsilon }_{1}}\,\,and\,\,\,{{\upsilon }_{2}}\] at A and B respectively. If C is the mid-point between A and B then velocity of the particle at C is equal to
\[\upsilon =\sqrt{\frac{\upsilon _{1}^{2}+\upsilon _{2}^{2}}{2}}\]
Sample problems based on uniform acceleration
Problem 40. A body A moves with a uniform acceleration \[a\] and zero initial velocity. Another body B, starts from the same point moves in the same direction with a constant velocity \[v\]. The two more...
Projectile Motion
A hunter aims his gun and fires a bullet directly towards a monkey sitting on a distant tree. If the monkey remains in his position, he will be safe but at the instant the bullet leaves the barrel of gun, if the monkey drops from the tree, the bullet will hit the monkey because the bullet will not follow the linear path.
The path of motion of a bullet will be parabolic and this motion of bullet is defined as projectile motion. If the force acting on a particle is oblique with initial velocity then the motion of particle is called projectile motion. Projectile. A body which is in flight through the atmosphere but is not being propelled by any fuel is called projectile. Example: (i) A bomb released from an aeroplane in level flight (ii) A bullet fired from a gun (iii) An arrow released from bow (iv) A Javelin thrown by an athlete Assumptions of Projectile Motion. (1) There is no resistance due to air. (2) The effect due to curvature of earth is negligible. (3) The effect due to rotation of earth is negligible. (4) For all points of the trajectory, the acceleration due to gravity ?g? is constant in magnitude and direction. Principles of Physical Independence of Motions. (1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other. This is called the principle of physical independence of motions. (2) The velocity of the particle can be resolved into two mutually perpendicular components. Horizontal component and vertical component. (3) The horizontal component remains unchanged throughout the flight. The force of gravity continuously affects the vertical component. (4) The horizontal motion is a uniform motion and the vertical motion is a uniformly accelerated retarded motion. Types of Projectile Motion. (1) Oblique projectile motion (2) Horizontal projectile motion (3) Projectile motion on an inclined plane
Oblique Projectile. In projectile motion, horizontal component of velocity (u cosq), acceleration (g) and mechanical energy remains constant while, speed, velocity, vertical component of velocity (u sin q), momentum, kinetic energy and potential energy all changes. Velocity, and KE are maximum at the point of projection while minimum (but not zero) at highest point. (1) Equation of trajectory: A projectile thrown with velocity u at an angle q with the horizontal. The velocity u can be resolved into two rectangular components.
v cos \[\theta \] component along X?axis and u sin q component along Y?axis. For horizontal motion \[x=u\text{ }cos\,\,\theta \times t\Rightarrow t=\frac{x}{u\cos \,\theta }\] ?. (i) For vertical motion \[y=(u\sin \,\theta \,)\,t-\frac{1}{2}\,g{{t}^{2}}\] ?. (ii) From equation (i) and (ii) \[y=u\sin \theta \left( \frac{x}{u\cos \,\theta } \right)\,-\frac{1}{2}g\,\left( \frac{{{x}^{2}}}{{{u}^{2}}{{\cos }^{2}}\,\theta } \right)\] \[y=x\,\tan \,\theta \,-\frac{1}{2}\frac{g{{x}^{2}}}{{{u}^{2}}{{\cos }^{2}}\,\theta }\] This equation shows more...
Aromatic compounds: These compounds consist of at least one benzene ring, i.e., a six-membered carbocyclic ring having alternate single and double bonds. Generally, these compounds have some fragrant odour and hence, named as aromatic (Greek word aroma meaning sweet smell).
These are also called benzenoid aromatics.
Note:q Non-benzenoid aromatics: There are aromatic compounds, which have structural units different from benzenoid type and are known as Non-benzenoid aromatics e.g. Tropolone, azulene etc.
(b) Heterocyclic compounds: Cyclic compounds containing one or more hetero atoms (e.g. O, N, S etc.) in the ring are called heterocyclic compounds. These are of two types:
Alicyclic heterocyclic compounds: Heterocyclic compounds which resemble aliphatic compounds in their properties are called Alicyclic heterocyclic compounds. For example,
Aromatic heterocyclic compounds: Heterocyclic compounds which resemble benzene and other aromatic compounds in most of their properties are called Aromatic heterocyclic compounds. For example,
(2) Classification based on functional groups: A functional group is an atom or group of atoms in a molecule that gives the molecule its characteristic chemical properties. more... 
Then distance travelled by particle \[=AB+BC=7\,m\]
(ii) Distance is a scalar quantity.
(iii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]\]
(iv) Unit: metre (S.I.)
(2) Displacement: Displacement is the change in position vector i.e., A vector joining initial to final position.
(i) Displacement is a vector quantity
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]\]
(iii) Unit: metre (S.I.)
(iv) In the above figure the displacement of the particle \[\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\]
\[\Rightarrow \,\,\left| AC \right|=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}+2(AB)\,(BC)\,\cos \,{{90}^{o}}\,}=\,\,5\,m\]
(v) If \[{{\vec{S}}_{1}},\,{{\vec{S}}_{2}},\,{{\vec{S}}_{3}}\,........\,{{\vec{S}}_{n}}\] are the displacements of a body then the total (net) displacement is the vector sum of the individuals. \[\vec{S}={{\vec{S}}_{1}}+\,{{\vec{S}}_{2}}+\,{{\vec{S}}_{3}}+\,........\,+{{\vec{S}}_{n}}\]
(3) Comparison between distance and displacement:
(i) The magnitude of displacement is equal to minimum possible distance between two positions.
So \[distance\ge \left| Displacement \right|\].
(ii) For a moving particle distance can never be negative or zero while displacement can be.
(zero displacement means that body after motion has came back to initial position)
i.e., Distance > 0 but Displacement > = or < 0
(iii) For motion between two points displacement is single valued while distance depends on actual path and so can have many values.
(iv) For a moving particle distance can never decrease with time while displacement can. Decrease in displacement with time means body is moving towards the initial position.
(v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without change in direction.
(vi) If \[{{\vec{r}}_{A}}\,\,and\,\,{{\vec{r}}_{B}}\] are the position vectors of particle initially and finally.
Then displacement of the particle
\[{{\vec{r}}_{AB}}={{\vec{r}}_{B}}-{{\vec{r}}_{A}}\]
and s is the distance travelled if the particle has gone through the path APB.
Sample problems based on distance and displacement
Problem 1. A man goes 10 m towards North, then 20m towards east then displacement is
[KCET (Med.) 1999; JIPMER 1999; AFMC 2003]
(a) 22.5 m (b) 25 m (c) 25.5 m (d) 30 m
Solution: (a) If we take east as \[x-\] axis and north as \[y-\] axis, then displacement \[=20\,\hat{i}\,+\,10\hat{j}\]
So, magnitude of displacement \[=\sqrt{{{20}^{2}}+{{10}^{2}}}=10\sqrt{5}=\text{ }22.5\,\,m.\]
Problem 2. A body moves over one fourth of a circular arc in a circle of radius r. The magnitude of distance travelled and displacement will be respectively
(a) \[\frac{\pi r}{2},\,r\sqrt{2}\] (b) \[\frac{\pi r}{4},\,r\] (c) \[\pi r,\,\frac{r}{\sqrt{2}}\] (d) \[\pi r,\,r\]
Solution: (a) Let particle start from A, its position vector \[{{\vec{r}}_{O}}_{A}=r\hat{i}\]
After one quarter position vector \[{{\vec{r}}_{OB}}=r\,\hat{j}.\]
So displacement \[=r\hat{j}-r\hat{i}\]
Magnitude of displacement \[=r\sqrt{2}.\]
and distance = one fourth of circumference \[=\frac{2\pi r}{4}=\frac{\pi r}{2}\]
Problem 3. The displacement of the point of the wheel initially in contact with the ground, when more... 
(b) Non-uniform (variable) speed: In non-uniform speed particle covers unequal distances in equal intervals of time. In the given illustration motorcyclist travels 5m in 1st second, 8m in 2nd second, 10m in 3rd second, 4m in 4th second etc.
Therefore its speed is different for every time interval of one second. This means particle is moving with variable speed.
(c) Average speed: The average speed of a particle for a given ‘Interval of time’ is defined as the ratio of distance travelled to the time taken.
\[Average\text{ }speed=\frac{\text{Distance travelled}}{\text{Time taken}}\]; \[{{v}_{av}}=\frac{\Delta s}{\Delta t}\]
q Time average speed: When particle moves with different uniform speed \[{{\upsilon }_{1}},\,\,{{\upsilon }_{2}},\,\,{{\upsilon }_{3}}\]... etc. in different time intervals\[{{t}_{1}},\,\,{{t}_{2}},\,\,{{t}_{3}}\], ... etc. respectively, its average speed over the total time of journey is given as
\[{{v}_{av}}=\frac{\text{Total distance covered}}{\text{Total time elapsed}}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}=\frac{{{\upsilon }_{1}}{{t}_{1}}+{{\upsilon }_{2}}{{t}_{2}}+{{\upsilon }_{3}}{{t}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}\]
Special case: When particle moves with speed \[{{v}_{1}}\] upto half time of its total motion and in rest time it is moving with speed \[{{v}_{2}}\] then \[{{v}_{av}}=\frac{{{v}_{1}}+{{v}_{2}}}{2}\]
q Distance averaged speed: When a particle describes different distances \[{{d}_{1}},\,\,{{d}_{2}},\,\,{{d}_{3}}\], ...... with different time intervals \[{{t}_{1}},\,\,{{t}_{2}},\,\,{{t}_{3}}\], ...... with speeds \[{{v}_{1}},{{v}_{2}},{{v}_{3}}......\] respectively then the speed of particle averaged over the total distance can be given as
\[{{\upsilon }_{av}}=\frac{\text{Total distance covered}}{\text{Total time elapsed}}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+......}=\frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}+......}{\frac{{{d}_{1}}}{{{\upsilon }_{1}}}+\frac{{{d}_{2}}}{{{\upsilon }_{2}}}+\frac{{{d}_{3}}}{{{\upsilon }_{3}}}+......}\]
q When particle moves the first half of a distance at a speed of v1 and second half of the distance at speed v2 then
\[{{v}_{av}}=\frac{2{{v}_{1}}{{v}_{2}}}{{{v}_{1}}+{{v}_{2}}}\]
q When particle covers one-third distance at speed v1, next one third at speed v2 and last one third at speed v3, then
\[{{v}_{av}}=\frac{3\,{{v}_{1}}{{v}_{2}}{{v}_{3}}}{{{v}_{1}}{{v}_{2}}+{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{1}}}\]
(d) Instantaneous speed: It is the speed of a particle at particular instant. When we say “speed”, it usually means instantaneous speed.
The instantaneous speed is average speed for infinitesimally small time interval (i.e., \[\Delta t\to 0\]). Thus
Instantaneous speed \[v=\underset{\Delta t\to 0}{\mathop{\lim }}\,\,\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt}\]
(2) Velocity: Rate of change of position i.e. rate of displacement with time is called velocity.
(i) It is a scalar quantity having symbol \[v\].
(ii) Dimension: \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{1}} \right]\]
(iii) Unit: metre/second (S.I.), cm/second (C.G.S.)
(iv) Types
(a) Uniform velocity: A particle is said to have uniform velocity, if magnitudes as well as more... 
e.g. (a) In uniform circular motion \[\theta =90{}^\text{o}\] always
(b) In a projectile motion \[\theta \] is variable for every point of trajectory.
(vi) If a force \[\overrightarrow{F}\] acts on a particle of mass m, by Newton’s 2nd law, acceleration \[\vec{a}=\frac{{\vec{F}}}{m}\]
(vii) By definition \[\vec{a}=\frac{d\vec{v}}{dt}=\frac{{{d}^{2}}\vec{x}}{d{{t}^{2}}}\left[ \text{As}\,\,\vec{v}=\frac{d\vec{x}}{dt} \right]\]
i.e., if x is given as a function of time, second time derivative of displacement gives acceleration
(viii) If velocity is given as a function of position, then by chain rule \[a=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=v.\frac{d\upsilon }{dx}\,\left[ \text{as}\,\,v=\frac{dx}{dt} \right]\]
(ix) If a particle is accelerated for a time t1 by acceleration a1 and for time t2 by acceleration a2 then average acceleration is \[{{a}_{a\upsilon }}=\frac{{{a}_{1}}{{t}_{1}}+{{a}_{2}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\]
(x) If same force is applied on two bodies of different masses \[{{m}_{1}}\] and \[{{m}_{2}}\] separately then it produces accelerations \[{{a}_{1}}\,\,and\,\,{{a}_{2}}\] respectively. Now these bodies are attached together and form a combined system and same force is applied on that system so that a be the acceleration of the combined system, then
more... 
From triangle ABC \[\tan \theta =\frac{BC}{AC}=\frac{AD}{AC}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}}\] …. (ii)
By comparing (i) and (ii) \[\operatorname{Velocity} = tan\,\theta \]
\[v = tan\,\theta \]
It is clear that slope of position-time graph represents the velocity of the particle.
Various position – time graphs and their interpretation
Then Total distance \[=\,|{{A}_{1}}|\,+\,|{{A}_{2}}|\,+\,|{{A}_{3}}|\,\]
= Addition of modulus of different area. i.e. \[s\,=\,\int{|\upsilon |\,dt}\]
Total displacement \[={{A}_{1}}+{{A}_{2}}+{{A}_{3}}\]
= Addition of different area considering their sign. i.e. \[r\,=\,\int{\upsilon \,dt}\]
here \[{{\operatorname{A}}_{1}}\,\,and {{A}_{2}}\] are area of triangle 1 and 2 respectively and \[{{A}_{3}}\] is the area of trapezium.
Acceleration: Let AB is a velocity-time graph for any moving particle
As \[Acceleration\text{ }=\frac{\text{Change in velocity}}{\text{Time taken}}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}\] .… (i)
From triangle ABC, \[\tan \theta =\frac{BC}{AC}=\frac{AD}{AC}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}\] …. (ii)
By comparing (i) and (ii)
Acceleration (a) \[=\tan \theta \]
It is clear that slope of velocity-time graph represents the acceleration of the particle.
Various velocity – time graphs and their interpretation
u = 0 [As body starts from rest]
a = +g [As acceleration is in the direction of motion]
v = g t … (i)
\[h=\frac{1}{2}g{{t}^{2}}\] … (ii)
\[{{\upsilon }^{2}}=2gh\] … (iii)
\[{{h}_{n}}=\frac{g}{2}(2n-1)\] ... (iv)
(ii) Graph of distance velocity and acceleration with respect to time: time:
(iii) As \[\operatorname{h} = \left( 1/2 \right)\,g{{t}^{2}}, i.e., h\propto {{t}^{2}}\], distance covered in time t, 2t, 3t, etc., will be in the ratio of \[{{1}^{2}}:\text{ }{{2}^{2}}:\text{ }{{3}^{2}}\], i.e., square of integers.
(iv) The distance covered in the nth sec, \[{{h}_{n}}=\frac{1}{2}g\,(2n-1)\]
So distance covered in I, II, III sec, etc., will be in the ratio of \[1 : 3 : 5\], i.e., odd integers only.
(2) If a body is projected vertically downward with some initial velocity
Equation of motion: \[\upsilon =u+g\,t\]
\[h=ut+\frac{1}{2}g\,{{t}^{2}}\]
\[{{\upsilon }^{2}}={{u}^{2}}+2gh\]
\[{{h}_{n}}=u+\frac{g}{2}\,(2n-1)\]
(3) If a body is projected vertically upward
(i) Equation of motion: Taking initial position as origin and direction of motion (i.e., vertically up) as positive
\[\operatorname{a} = g\] [As acceleration is downwards while motion upwards]
So, if the body is projected with velocity u and after time t it reaches up to height h then
\[\upsilon =u-g\,t\]; \[h=ut-\frac{1}{2}g\,{{t}^{2}}\]; \[{{\upsilon }^{2}}={{u}^{2}}-2gh;\,\,{{h}_{n}}=u-\frac{g}{2}\,(2n-1)\]
(ii) For maximum height \[\operatorname{v} = 0\]
So from above equation
\[\operatorname{u} = gt,\]
\[h=\frac{1}{2}g{{t}^{2}}\]
and \[{{u}^{2}}=2gh\]
(iii) Graph of distance, velocity and acceleration with respect to time (for maximum height):
It is clear that both quantities do not depend upon the mass of the body or we can say that in absence of air resistance, all bodies fall on the surface of the earth with the same rate.
(4) In case of motion under gravity for a given body, mass, acceleration, and mechanical energy remain constant while speed,
The path of motion of a bullet will be parabolic and this motion of bullet is defined as projectile motion. If the force acting on a particle is oblique with initial velocity then the motion of particle is called projectile motion. Projectile. A body which is in flight through the atmosphere but is not being propelled by any fuel is called projectile. Example: (i) A bomb released from an aeroplane in level flight (ii) A bullet fired from a gun (iii) An arrow released from bow (iv) A Javelin thrown by an athlete Assumptions of Projectile Motion. (1) There is no resistance due to air. (2) The effect due to curvature of earth is negligible. (3) The effect due to rotation of earth is negligible. (4) For all points of the trajectory, the acceleration due to gravity ?g? is constant in magnitude and direction. Principles of Physical Independence of Motions. (1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other. This is called the principle of physical independence of motions. (2) The velocity of the particle can be resolved into two mutually perpendicular components. Horizontal component and vertical component. (3) The horizontal component remains unchanged throughout the flight. The force of gravity continuously affects the vertical component. (4) The horizontal motion is a uniform motion and the vertical motion is a uniformly accelerated retarded motion. Types of Projectile Motion. (1) Oblique projectile motion (2) Horizontal projectile motion (3) Projectile motion on an inclined plane
Oblique Projectile. In projectile motion, horizontal component of velocity (u cosq), acceleration (g) and mechanical energy remains constant while, speed, velocity, vertical component of velocity (u sin q), momentum, kinetic energy and potential energy all changes. Velocity, and KE are maximum at the point of projection while minimum (but not zero) at highest point. (1) Equation of trajectory: A projectile thrown with velocity u at an angle q with the horizontal. The velocity u can be resolved into two rectangular components.
v cos \[\theta \] component along X?axis and u sin q component along Y?axis. For horizontal motion \[x=u\text{ }cos\,\,\theta \times t\Rightarrow t=\frac{x}{u\cos \,\theta }\] ?. (i) For vertical motion \[y=(u\sin \,\theta \,)\,t-\frac{1}{2}\,g{{t}^{2}}\] ?. (ii) From equation (i) and (ii) \[y=u\sin \theta \left( \frac{x}{u\cos \,\theta } \right)\,-\frac{1}{2}g\,\left( \frac{{{x}^{2}}}{{{u}^{2}}{{\cos }^{2}}\,\theta } \right)\] \[y=x\,\tan \,\theta \,-\frac{1}{2}\frac{g{{x}^{2}}}{{{u}^{2}}{{\cos }^{2}}\,\theta }\] This equation shows
