# 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 displacement$s_2$ 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 $v_1$and 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-1/2$gt^2$  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+1/2$gt^2$. (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 g$t_1t_2$   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 $90^0$)

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