Speed of a partical as it approaches stagnation point

[Wildcat04]

Homework Statement

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)

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?

 

[rl.bhat]

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

 

[tomcue]

I would approach this problem by first understanding the physical meaning behind the equations and variables given. The first equation, s = 0.6ft upstream @ t=0, represents the position of a particle in a fluid flow at a specific time. The second equation, s = 0.6 e^(-0.5 t), represents the position of the same particle as a function of time. This equation can be obtained by solving the differential equation for the position of a particle in a fluid flow, which is described by the Navier-Stokes equation.

a) To determine the speed of the particle as a function of time, we can take the derivative of the position function with respect to time, since speed is the rate of change of position with respect to time. This gives us V(t) = -0.3 e^(-0.5t) ft/s, as you correctly calculated. This means that the speed of the particle decreases over time, as seen by the negative sign in front of the exponential function.

b) To determine the speed of the fluid as a function of position, we need to understand that the fluid velocity is related to the position of the particle in the flow. As the particle moves along the streamline, the fluid moves with it. This means that the speed of the fluid at a certain position is equal to the speed of the particle at that same position. Therefore, V(s) = V(t) = -0.3 e^(-0.5t) ft/s. Since we are given the position of the particle as a function of time, we can substitute t = t(s) into the equation to get V(s) = -0.3 e^(-0.5t(s)) ft/s. This is the same as the solution given by your instructor, -0.5s ft/s.

c) To determine the fluid acceleration along the streamline, we can use the material derivative, which takes into account the change in velocity of the fluid as the particle moves along the streamline. The material derivative is given by dV/dt + u dv/ds, where u is the fluid velocity and dv/ds is the rate of change of velocity with respect to position. Using the solution for V(s) from part b, we can calculate the material derivative as DV/DT = -0.3 (-0.5 e^(-0.5t(s))) + u (-0.5