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Asked by Prashant DIGHE 26th April 2020, 10:09 PM
Answered by Expert
Answer:
velocity vector , begin mathsize 14px style v with rightwards arrow on top space equals space k subscript 1 space i with hat on top space plus space k subscript 2 space x space j with hat on top space end style
x-component of velocity ,  ( dx/dt )  = k1 ................(1)
 
If we integrate eqn.(1) by considering x = 0 at t = 0, then we get  x = k1 t   .................... (2)
 
y-component of velocity  ( dy/dt ) = k2 x  = k2 k1 t   ...............(3)
 
In above equation x is substituted using eqn.(2)
 
By integrating eqn.(3) by considering y = 0 at t = 0, we get y = (1/2)( k2 k1 ) t2  ...................(4) 
 
Equation (2) and (4) gives the coordinate of particle as a function of time.
 
Trajectory of particle is obtained from eqn. (2) and eqn.(4) by eliminating time t
 
let us substitute t = x / k1 using eqn.(2) and rewrite eqn.(4) as ,
 
y = (1/2) ( k2 k1 ) [ x/k1 ]2 = [ k2 / ( 2 k1 ) ] x2
Answered by Expert 27th April 2020, 8:27 AM
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