when  the body changes it’s position  with respect to reference point and time than body is said to be in motion. Motion can be of  different type depends upon the path adopted i) Circulatory Motion  which have a circular path  ii) Linear motion  when body moves in straight line   iii) Oscillatory/ vibratory motion  when a body moves in to and froth path.

PHYSICAL QUANTITY  There  are seven  basic physical quantities ie length, time, mass,  temperature , amount of substance , electric current and luminous intensity  they have both magnitude and unit but they are classified in two groups a) Scalar quantity the physical quantity having only magnitude such as length, mass, time  b) vector quantity is that physical quantity which have both magnitude and direction such as displacement , velocity , force  weight


Distance is the length of path travelled by the body  whereas displacement is the shortest distance between initial and final position of the body


example 1

Example 2


( examples  taken from support material NCT )

UNIFOM  MOTION : When a body travels equal distance in equal interval of time  example movement of hands of the  clock .or a person travel  2 meter in first second, again 2 meters in next  second and so  on .

NON UNIFORM  MOTION : When a body travels unequal distance in equal intervals of time such as  in first  a person walks 3 meters than in second he walks 4 meters than in third second he travels 2 meters

SPEED : The distance travelled by body in unit time is called speed. It is scalar quantity unit  SI  meter/ second or Km/ hr  Speed = distance/ time If body is  moving in uniform  motion than the speed is constant but in non uniform  motion the speed is not same average speed   can be calculated as  Total distance travelled in journey divided by total time taken in journey. Average Speed = Total Distance Travelled / Total time taken

Example : What will be the speed of the body  in m/s and km/hr if body travel 60 kms in 6hrs

speed = distance /time = 60/6= 10 km/hr

in m/s (multiply km/hr  by 5/18  = speed in m/s)   so  10×5/18= 50/18=25/9 m/sec

VELOCITY : The speed of the moving body in given specific direction is called velocity . It is a vector quantity Its value changes either magnitude or direction changes SI unit meter/ second.

AVERAGE VELOCITY  It is the mean value of initial velocity (u) and final velocity (v)

Average velocity = (u + v) /2

Example : During  first half  of a journey  a body travels with the speed of 40 km/hr  and next half it travels with the speed of 20km/hr than the  average speed  is

40+20 = 60 >>> 60/2= 30 km/hr

Example if a car travels  20km in first hour  60 km in second hour and 40 km in the third hour  than  average speed =( 20+60+40) -:- 3= 120 km /3 = 40 km/hr

ACCELERATION : It is the rate of change of velocity is called acceleration .

Acceleration = (Final velocity – initial velocity)- -: –_ Time  In this case v>u

a= (v-u) / t     SI unit meter/ second^2

RETARDATION/ DEACCELERATION : The rate of decrease in velocity is called deacceleration  that is change in velocity -:- time  it is also called negative velocity SI unit meter/ sec^2 In this case u>v

Examples : a) A car is travelling with the speed  of 40 km/hr increases to 70 km/hr in 5 sec than  acceleration  of car = change in speed = 70-40 = 30 km/hr

change in speed in m/s = 30 x5/18  = 5×5/3= 25/3  acceleration = change in speed -;- time

25/3-;- 5 = 5/3 meter/ sec^2

example b) If a car is moving  with the speed  of 20 km/hr comes to  rest  in .5 hr than the retardation is  = 0-20 = -20 km/hr  now

time taken  ‘5 hr   than deacceleration = change in speed -;- time = -20 -;- .5 = 40 km/hr^2At constant velocity  acceleration is Zero

Velocity increases  create the accelerate a=( v2- v1) -:-  (t2-t1)

ii) Body A has more velocity as slope is more


This graph represent deacceleration


 equation of motionLet  us suppose  a body starts moving with the velocity u  at A  after time of t seconds it’s velocity becomes v at B  . Now by graph  change in velocity is BC -DC = BD====> V-U                           time taken  = OC  = t

Acceleration = change in velocity / time =====> a = (V-U)/t===> at = V-U ====> V= U+ at

first equation of motion is ==== >  v= u +at

Total distance travelled  (S)   as we know that average velocity  =( Initial velocity + Final velocity )/2

total distance travelled = average velocity x time = => equal to aera of the given figure  OABDCO which is trapezium whose aera = 1/2x( sum of parallel sides ) height ==> sum of parallel sides = OA+BC =(u+v  ) height=OC= t

S= 1/2 x (v+u ) t  ====> put the value of  v=u+at       S= 1/2x(u+at+ u) = 1/2 (2u + at)t

S= ut + 1/2 at^2  (second equation of motion )

as we know that   distance travelled= area of given figure OABDCO =  S= (V+U) t/2  put the value of t from equation  first ===> v= u+at ==> t=( v-u)/a

S = (v+ u) (v-u)/2a ====>  2aS = (v+u) (v-u)====>   2as = V^2 -U^2 ==> U^2 + 2aS = V^2

v^2 = U^2 + 2aS ( third equation of motion )

Example a) A car  starting from rest  moves with uniform speed with acceleration of 0.2 m/s^2 foe 4 minutes than the speed  (1st equation )     v= u+at  v=?  u=0  t = 4×60 = 240 sec than V=0+0.2×240 = 48 m/s and  distance travelled S = ut+ 1/2at^2  = 0x240 + 1/2 x0.2×240 = 0 +24= 24 meters

b)When is applied to car it produces deacceleration of 6m/s^2  opposite to the direction of motion if it stops in 2 sec than  the distance travelled will be  S= ut + 1/2 at^2  first we have to find u =? as body stops therefore v=0

V= U- at ==> 0 = U – 6×2  ==>  u= 12  now in the  second equation  s=12×2 – 1/2x6x2x2 = 24-12 = 12 meter

UNIFORM CIRCULAR MOTION If the bodyis moving in circular path with uniform speed it’s motion is called uniform circular motion. In such kind of motion speed reamin same  but direction changes every moment  so it produce an acceleration during the circular motion  velocity = 2xn r/t  (n= pie = 2/7)

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