Important formulas of Kinematics

    • Average Speed = frac{Totaldistancetraveled}{Total time taken}
    • If the body covers 1st half of distance with a speed x and the second half with a speed y,then the average speed = frac{2xy}{x+y}
    • 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 =frac{3xyz}{xy+yz+zx}
    • Average velocity = frac{TotalDisplacement}{Total time taken}
    • If a body travels a displacement s_1 in t_1 seconds and a displacement s_2 in t_2 seconds, in the same direction then   Average velocity = frac{s_1+s_2}{t_1+t_2} .
    • If a body travels a displacement s_1 with velocity v_1 , and displacements_2 with velocity v_2 in the same direction then Average velocity = frac{( s_1 + s_2) v_1v_2}{s_1v_2+s_2v_1}
    • If a body travels first half of the displacement with a velocity v_1and next half of the displacement with a velocity v_2 in the same direction , then                                                          Average velocity = frac{2v_1v_2}{v_1+v_2} .
    • If a body travels a time t_1 with velocity v_1 and for a time t_2 with a velocity v_2 in the same direction then Average velocity = frac{v_1t_2+V_2t_1}{t_1+t_2} .
    • If the body travels 1st half of the time with a velocity v_1 next half of the time with a velocity v_2 in same direction , then  Average velocity = frac{v_1+v_2}{2}
    • 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+frac{1}{2}at^2 c) v^2u^2 =2as
    • Distance traveled  in the nth second s_n=u+a(n-1/2)
    • 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 gt^2 c) v^2 = 2gs and the equation for the distance traveled in nth second changes to s_n=g(n-1/2)
    • Equations of motion of a body projected up vertically :(we will obtain these equations by substitution a=-g in equations of motion)
    • a) v=u-gt  b) S=ut-1/2gt^2 c) v^2-u^2=-2gs and s_n=u-g(n-1/2)
    • Equation for maximum height reached H_{max} = frac{u^2}{2g}RightarrowH_{max} alpha u^2
    • Time of ascent t_a=frac{u}{g}t_aalpha u
    • Time of descent t_d =frac{u}{g}t_dalpha 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+1/2gt^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=sqrt(2gh)
    • If the displacements of a body in m^{th} ,n^{th} seconds of its journey.Then the uniform acceleration of the body a=frac{s_n-s_m}{n-m}
    • From the above equation we can observe that by substituting n=1,2,3,4,…. we get a=s_2-s_1=s_3-s_2=……….. =s_{n-s}s_{n-1} =a.
    • A body projected up with velocity u from the top  of a tower reaches ground in t_1 seconds.If  it  is thrown down with the same velocity u it reaches ground in t_2 seconds.Then, when it is dropped freely the time taken to reach the ground will be t=sqrt(t_1t_2) and h=1/2 gt_1t_2 and t_1 -t_2 =2 frac{u}{g}.
    • Projectile motion :Let us suppose that a projectile is projected with an initial velocity u making an angle theta with x axis. a)Horizontal component of velocity u_x = u costheta , and u_x =v_x ,which will be constant through out the flight of the projectile as horizontal component of acceleration a_x = 0.  b)Vertical component of velocity of the projectile u_y = u sintheta. Vertical component of velocity at any time of its journey v_y =u_y-gt or v_y =usintheta -gt.   c)Magnitude of the resultant velocity V = sqrt(v_x^2+v_y^2) and the angle x made by v with the horizontal is given by Tanalpha= frac{v_y}{v_x}
    • Time of ascent = frac{usintheta}{g}
    • Time of descent = frac{usintheta}{g}
    • Time of flight = frac{2u sintheta}{g}
    • Maximum height reached H_{max} =frac{u^2 sin^2theta}{2g} ;frac{H1}{H2} =frac{sin^2theta_1}{sin^2theta_2} when u is same
    • Horizontal Range R= frac{(u^2sin2theta)}{g }=frac{2u^2 sintheta costheta}{g} a)R is maximum when theta =45^0 b)R_{max} =frac{u^2}{g} c) If T is the time of flight, R=u costhetatimes t d)For given velocity of projection R is same for the angles of projections theta and (90-Theta) Ex: 25 and 65 i.e the Range for those two angles will be same whose sum is 90^0)