- #1

vdfortd

- 10

- 0

## Homework Statement

An engineer designing a system to control router for a machining process models the system so that the router's acceleration (in in/s

^{2}) during an interval of time is given by a= -0.4v, where v is the velocity of the router in in/s

^{2}. When t=0, the position is s=0 and the velocity is v= 2 in/s. What is the position at t=3 seconds?

## Homework Equations

a= dv/dt

## The Attempt at a Solution

This section has to do with straight line motion when the acceleration depends on velocity or position.

This is what I tried to do:

a= -0.4v

a= dv/dt

dv/dt = -0.4v

dv/v= -0.4 dt

I integrated both sides and I got:

ln v = -.4t + C ( C = constant)

From my differential equations math class, we did problems like these and we got rid of the ln by making it all a power of e.

e

^{ln v}= e

^{ -.4t + C}

you get v= e

^{-.4t+C}or v= e

^{-.4t}C

Now I know this is wrong because we know that when t=0 v=2. Am I on the right track or did i do this completely wrong? Thanks for any help.