- #1
bobsagget
- 18
- 0
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 because v1=0
then i subbed d=29m-d(K) and got
29-D(of kirk) = 1/2(acc)(time^2)
i don't know what to do after? any help? or if I am doing it right at all please help!
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 because v1=0
then i subbed d=29m-d(K) and got
29-D(of kirk) = 1/2(acc)(time^2)
i don't know what to do after? any help? or if I am doing it right at all please help!