# Speed of a partical as it approaches stagnation point

1. Sep 14, 2009

### Wildcat04

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?

2. Sep 14, 2009

### rl.bhat

v = -0.5*0.6e^(-0.5t)
= -0.5*S
Now find dv/ds.