Kinematics :

 Average Speed =
 If the body covers 1st half of distance with a speed x and the second half with a speed y,then the average speed =
 If the body covers 1st 1/3rd of a distance with a speed x , and 2nd 1/3 with a speed y , and the 3rd 1/3rd distance with a speed z, then average speed =
 Average velocity =
 If a body travels a displacement in seconds and a displacement in seconds, in the same direction then Average velocity = .
 If a body travels a displacement with velocity , and displacement with velocity in the same direction then Average velocity =
 If a body travels first half of the displacement with a velocity and next half of the displacement with a velocity in the same direction , then Average velocity = .
 If a body travels a time with velocity and for a time with a velocity in the same direction then Average velocity = .
 If the body travels 1st half of the time with a velocity next half of the time with a velocity in same direction , then Average velocity =
 For a body moving with uniform acceleration if the velocity changes from u to v in t seconds, then Average velocity = (u+v)/2 .
 Equations of motion of a body moving with uniform acceleration along straight line.
 a) V=u+at b) S=ut+ c) – =2as
 Distance traveled in the nth second =u+a(n1/2)
 Equations of motion for a freely falling body ( Note: we can obtain these equations by substitution of u=0 and a=g in above equations . a) v=gt b) S=1/2 c) = 2gs and the equation for the distance traveled in nth second changes to =g(n1/2)
 Equations of motion of a body projected up vertically :(we will obtain these equations by substitution a=g in equations of motion)
 a) v=ugt b) S=ut1/2 c) =2gs and =ug(n1/2)
 Equation for maximum height reached =
 Time of ascent =; u
 Time of descent =; u
 Time of flight T=2u/g
 When a body is thrown up from top of a tower or released from a rising baloon,with velocity u.Displacement traveled before reaching ground S=ut+1/2. (t= time during which the object is in the air and S=h=height of the tower).
 When a body is dropped from a tower of height h and another body is thrown up vertically with a velocity u then they will meet after t=h/u seconds.
 When a body is dropped from a tower of height h . Its velocity when it reaches ground v=
 If the displacements of a body in seconds of its journey.Then the uniform acceleration of the body a=
 From the above equation we can observe that by substituting n=1,2,3,4,…. we get a===……….. = =a.
 A body projected up with velocity u from the top of a tower reaches ground in seconds.If it is thrown down with the same velocity u it reaches ground in seconds.Then, when it is dropped freely the time taken to reach the ground will be t= and h=1/2 g and =2 .
 Projectile motion :Let us suppose that a projectile is projected with an initial velocity u making an angle with x axis. a)Horizontal component of velocity , and ,which will be constant through out the flight of the projectile as horizontal component of acceleration = 0. b)Vertical component of velocity of the projectile . Vertical component of velocity at any time of its journey =gt or = gt. c)Magnitude of the resultant velocity V = and the angle x made by v with the horizontal is given by =
 Time of ascent =
 Time of descent =
 Time of flight =
 Maximum height reached = ; = when u is same
 Horizontal Range R= = a)R is maximum when = b) = c) If T is the time of flight, R= d)For given velocity of projection R is same for the angles of projections and Ex: 25 and 65 i.e the Range for those two angles will be same whose sum is )
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Equation aren’t clear.Pl. Clarify
Equation aren’t clear.Pl. Clarify and explain
nice.thanks!
Hi, thanks for these information
Helped me a lot !! thanks :)
coooool
it would better if u had explain it in better way other way this was great
so good
thank u for helping