2 PARTICLES OF A AND B START MVING SIMULTANEOUSLY ALONG THE LINE JOINING THEM IN THE SAME DIRECTION WITH ACC OF 1M/S^2 AND 2 M/S^2 AND SPEEDS 3 M/S AND 1 M/S RESPECTIVELY.

INITIALLY A IS 10m BEHIND B. WHAT IS THERE MINIMUM DISTANCE BETWEEN THEN.

IN D.C. PANDEY IT IS WRITTEN TO TAND ASSUME ONE OF THE BODIES TO BE AT REST  TAKE OUT THE RELATIVE VELOCITY AND REL ACC AND THE DISTANCE BETWEEN THE RELATIVE PATH TAKEN AND THE DISTANCE BETWEEN THE BODY AT REST

BUT IT IS NOT BEING SOLVED BY ME

PLS REPLY

Ans is 8m.Using the method you mentioned,

Let us assume B to be at rest.Then,

Rel.vel of A w.r.t B=V_a-V_b=2m/s

And rel.acc of A w.r.t B=a_a-a_b=-1m/s²

Since the acceleration is -ve,A will come to rest after some time in this frame of reference

So,V²-U²=2aS-4=-2SS=2m

Hence the min. distance between them=8m

Alter method:

S_a=3t+(1/2)t² and S_b=t+t² after t sec.So,the distance between them after t sec is d=10+S_b-S_a=10+(1/2)t²-2t which is to be minimised.

Using calculus,critical point is t=2s and hence,min.dis=10+2-4=8m