Q: Three particles starts from rest simaltaneously along three smooth chords of a vertical circle joining at the lowest point of the circle . If the inclination of the chord is 30 degree, 60 degree , 90 degree respectively. Find the ratio of the time taken by them to move down the chordsQ: the velocity 'v' and displacement 's' a body is moving in a 1-D is given by v2 = ks where k is a const. if displacement at t=0 is zero.The velocity after 2 sec will be

Ans :: let the time taken by three particles on chord 90⁰ , 60⁰, and 30⁰ are t₁, t₂ and t₃ and let the radius of circle is R

initial velocity of particles are 0

therefore, 1/2(g)(t₁)² = 2*R                           eq (1)

similarly 1/2(g)(t₂)² = 2*R*cos30⁰                           eq (2)

and 1/2(g)(t₃)² = 2*R*cos60⁰                           eq (3)

from eq (1), eq (2) and eq (3), we get

t₁ : t₂ : t₃   = (2)^1/2 : 1: (3)^1/4

Ans:: since intial displacement and velocity are zero

therefore, s =1/2 gt² and v² = 2as,  where a is the acceleration

so comparing with v²=ks we get new eqs as

s =1/4kt²
therefore after 2 sec, s= 1/4k(2)²

s=k

therefore v = k^1/2