# Homework Help: Acceleration, Velocity and distance

1. Nov 18, 2008

### Oblivion77

1. The problem statement, all variables and given/known data

An engineer designing a system to control a router for a machining process models the system so that the router's acceleration during an interval of time is a=-1.8vm/s . When t=0 , its position is s=0 and its velocity is 2.4 m/s. Determine the router's position (unit: m) at s.

2. Relevant equations

a=dv/dt, v=ds/dt

3. The attempt at a solution

First i did -1.8v=dv/dt did some integration and work and found v = e^(-1.8t)

I then used e^(-1.8t) = ds/dt and solved for s and subsituted t=0(lower bound) and t=0.75(upper bound) and found s = 0.412m

I am not sure if the first step I did is allowed since I never used the v=2.4. Please tell me my mistakes. Thanks!

2. Nov 19, 2008

### tiny-tim

Hi Oblivion77!

You integrated twice, but you only put in a constant of integration the second time.

The v = 2.4 comes in the first time!

3. Nov 19, 2008

### Oblivion77

Ok, using the v=2.4 with the first integration I get -1.8v=dv/dt, doing some integration I get
t=(-1/1.8)ln(v)+0.49 (does this look right so far?). Now I want to solve for v and use v=ds/dt to get the function for position. so v=e^(0.88-1.8t), e^(0.88-1.8t)=ds/dt .Then to sub t=0.75s for the upper bound and t=0 for the lower bound. Does this look right now? Or did I make a mistake somewhere? Thanks!

4. Nov 19, 2008

### tiny-tim

oooh

No, you integrated dv/v - -1.8dt to get v = Ce-1.8t.

And so, when t= 0, e-1.8t = … ? , and so C = … ?

5. Nov 20, 2008

### Oblivion77

Ok, so when i do that I get v=Ce^(-1.8t). And the intial conditions are t=0, v=2.4. So would i do 2.4=Ce^0, C=2.4. Then v=2.4e^(-1.8t), or do i not use the velocity?

6. Nov 20, 2008

### tiny-tim

What a long question!

Yes, v = 2.4e-1.8t

why does that worry you?