1. Oct 13, 2009

### bobsagget

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!

2. Oct 13, 2009

### and9

Good set up. You have the right distance formulas for both. Solve andrew's distance formula in terms of D then set both distance formulas equal to each other and you should find a time t that they intersect or meet up.

(You will find that Andrew's distance = 29 - 1/2at^2)

3. Oct 13, 2009

### bobsagget

oooo ok thanks, now i understand the rest