Saturday, January 3, 2009

Important Formulas in Physics

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^2 - u^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} \Rightarrow H_{max} \alpha u^2
    • Time of ascent t_a=\frac{u}{g}; t_a\alpha u
    • Time of descent t_d =\frac{u}{g}; t_d \alpha 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 cos\theta , 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 sin\theta. Vertical component of velocity at any time of its journey v_y =u_y-gt or v_y =usin\theta -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 Tan\alpha= \frac{v_y}{v_x}
    • Time of ascent = \frac{usin\theta}{g}
    • Time of descent = \frac{usin\theta}{g}
    • Time of flight = \frac{2u sin\theta}{g}
    • Maximum height reached H_{max} =\frac{u^2 sin^2\theta}{2g} ;\frac{H1}{H2} =\frac{sin^2\theta_1}{sin^2\theta_2} when u is same
    • Horizontal Range R= \frac{(u^2sin2\theta)}{g }=\frac{2u^2 sin\theta cos\theta}{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 cos\theta\times 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)
Posted by chavan gyanraj at 12:45 AM
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