A particle `A` moves with velocity `(2hati-3hatj)m//s` from a point `(4,5m)m`. At the same instant a particle `B`, moving in the same plane with velocity` (4hati+hatj)m//s` passes through a point `C(0,-3)m`. Find the `x`-coordinate (in `m`) of the point where the particles collide.

Answer

8

Explanation

If particle collide sfter time t at point P, then position of both after time t should be same both have uniform motion.
by `S=S_(0)+vxxt`
for `I^(st)` particle `S_(I)=4hat(i)+5hat(j)+(2hat(i)-3hat(j))xxt`
for `II^(nd)` particle `S_(II)=-3hat(j)+(4hat(i)+hat(j))xxt`
but `S_(I)=S_(II)`
`4hat(i)+5hat(j)+(2hat(i)-3hat(j))t=-3+(4hat(i)+hat(j))+(4hat(i)+hat(j))xxt`
`4hat(i)+8hat(j)=2hat(i)t+4hat(i)t`
by comrision along `xxDeltay`
`t=2` sec
`S_(I)=4hat(i)+5hat(j)+(2hat(i)-3hat(j))2`
`=8hat(i)-hat(j)`
`x=8 m`