1. The problem statement, all variables and given/known data s = 0.6ft upstream @ t=0 s = 0.6 e^(-0.5 t) where t is in seconds and s is in ft a) Determine the speed of the particle as a function of time Vpart(t) b) Determind the speed of the of the fluid as a function of the position along the streamline V=V(s) c) Determine the fluid acceleration along the streamline as a function of position as = as(s) (solutions provided by instructor) 3. The attempt at a solution a) Take the derivative of s with respect to t (simple enough) V(t) = -0.3 e^(-0.5t) ft/s <-correct solution b) solution = -0.5s ft/s So I thought that I would solve the first given equation for t and differentiate, however I was told that all I needed to do was put in t = t(s) into (a). My question is how does this work and how do you come up with -0.5s. I am sure it is simple and right in front of me, but I can't get my head around it. c) solution = 0.25s Material derivative of (b) (using the correct answer) DV/DT = dV/dt + u dv/ds = 0 + (-.5)*(-.5s) = 0.25s <- Correct solution Can anyone help be clear up part b?