CBSE Class 11-science Answered
explain all the derivations of chapter-motion in a plane
Asked by ABHILASHA | 16 Sep, 2019, 06:56: PM
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, 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 :-
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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