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CBSE Class 11-science Answered

explain all the derivations of chapter-motion in a plane
Asked by ABHILASHA | 16 Sep, 2019, 06:56: PM
answered-by-expert Expert Answer
Let us consider x-y coordinate system in a plane and an object is moving in an arbitrary direcion in that plane.
 
If we know (x,y) coordinate of the object as a function of time, we can determine vx and vy
which are x-component and y-component of velocity respectively.
 
Then velocity v of the object is given by,   begin mathsize 14px style v space equals space square root of v subscript x superscript 2 space plus space v subscript y superscript 2 end root space end style  and the direction is given by  tanθ = ( vy / vx
or  θ = tan-1( y/x )  , where θ is angle made by velocity direction with x-axis
 
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if an object moves in plane in any arbitrary direction with constant acceleration a,
its position vector r at any time t is given by,
 
r = r0 + v0t + (1/2) a t2 ..........................(1)
 
where r is position vector of the object at time t, r0 is position vector of the object at initial time t = 0,
v0 is the velocity and a is acceleration
 
Above equation can be written as component form as
 
x = x0 + v0x t + (1/2) ax t2 
 
y = y0 + v0y t + (1/2) ay t2 
 
where x and y are x-coordinate and y coordinate respectively at time t .  suffix 0 stands for initial values, 
x stands for x-component quantities and y stands for y-component quantities
 
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projectile motion
 
Let u0 be the initial projection velocity that makes angle θ0 with horizontal.
 
In component form,  horizontal component u0x = u0 cosθ0   and  vertical component u0y = u0 sinθ0
 
x-position at any time t :-   x = ( u0 cosθ0) t  ;  y-position at any time t :-  y = ( u0 sinθ0) t - (1/2) g t2 
 
velocity components at any time t :-   ux = u0 cosθ0  ,    uy = ( u0 sinθ0) - g t
 
direction of resultant velocity :-  tanθ = ( uy / ux ) ,  where θ = tan-1( uy / ux
where θ is angle made by velocity direction with x-axis
 
Equation of projectile :-  begin mathsize 12px style y space equals space open parentheses tan theta subscript 0 space close parentheses space x space minus space fraction numerator g space x squared over denominator 2 space open parentheses v subscript 0 cos theta subscript 0 close parentheses squared end fraction space end style
 Maximum height  hm = ( u0 sinθ0 )2 / (2g )
 
Horizontal range R =  u02 sin2θ0  / g  ;  maximum range is when θ0 = 45° , Rmax = u02 / g
 
time to reach maximum height ,  t = u0 sinθ0  / g   ;  
 
Time of flight , i.e., staring from projection to reaching the ground ,  Tf = ( 2 u0 sinθ0 ) / g
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Circular motion
 
When a body is moving in a circular path with constant speed v,
then it is subjected to acceleration a directed towards the centre of circular path which is known as centripetal acceleration.
 
a = v2 / R   = ω2 R
 
where ω is angular velocity defined as  v = ω R    and R is radius of circular path
 
 
 
Answered by Thiyagarajan K | 19 Sep, 2019, 11:19: AM
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