three particles start moving simultaneously from a point on a horizontal smooth plane. 1st particle moves with velocity v1 towards east, 2nd with velocity v2 towards north and 3rd one towards north east. find the velocity of the 3rd particle so that the three always lie on a straight line. (in terms of v1 & v2)
three particles start moving simultaneously from a point on a horizontal smooth plane. 1st particle moves with velocity v1 towards east, 2nd with velocity v2 towards north and 3rd one towards north east. find the velocity of the 3rd particle so that the three always lie on a straight line. (in terms of v1 & v2)
Let us consider particles starts from origin, positive x and positive y are taken east and north respectively.
The position of first particle at time t is (v1t,0)
The position of second particle at time t is (0,v2t)
The equation of line joining these two positions
y = [(v2t - 0) / (0-v1t)] (x-v1t) + 0
y = -(v2/v1)(x-v1t)
=> y = -(v2/v1)x + v2t.......(1)
Equation of the path of third particle
y = x .......(2)
from 1 & 2 we get,
x = y = [v1v2/(v1+v2)]t
distance of this point from origin d = (root 2)[v1v2/(v1+v2)]t
If third particle attains such distance in time t then the will be lie on a line
therefore
v3 t = d
v3 = root 2[v1v2/(v1+v2)].
Let us consider particles starts from origin, positive x and positive y are taken east and north respectively.
The position of first particle at time t is (v1t,0)
The position of second particle at time t is (0,v2t)
The equation of line joining these two positions
y = [(v2t - 0) / (0-v1t)] (x-v1t) + 0
y = -(v2/v1)(x-v1t)
=> y = -(v2/v1)x + v2t.......(1)
Equation of the path of third particle
y = x .......(2)
from 1 & 2 we get,
x = y = [v1v2/(v1+v2)]t
distance of this point from origin d = (root 2)[v1v2/(v1+v2)]t
If third particle attains such distance in time t then the will be lie on a line
therefore
v3 t = d
v3 = root 2[v1v2/(v1+v2)].
