Kirk and andrew are at opposite ends of a hallway stretching 29.0 m. Andrew accelerates from REST towards kirk at a constant rate of 0.18m/s^2. K irk walks towards andrew at a constant veloctiy of 3.0m/s. HOw much time elaspes before kirk and andrew high five". kirk v1= 3.0 m/s acc= 0 m/s^2 distance (K)=? time passed =? andrew v1= 0 m/s time passed =? acc= 0.18 m/s^2 distance= 29m -d(k) i then got an equation for kirk which was delta d= Vavg/ time and andrews was d=v1(delta)T+1/2(acc)(time^2) ^ ^ ^ cancels cuz v1=0 then i subbed d=29m-d(K) and got 29-D(of kirk) = 1/2(acc)(time^2) i dont know what to do after??? any help????????? or if im doing it right at all please help!