Solve Position at t=3s for Motion Homework: Dynamics Help

[vdfortd]

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/s²) during an interval of time is given by a= -0.4v, where v is the velocity of the router in in/s². 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^-.4tC

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.

 

[nvn]

vdfortd: Nice work. You are doing well, so far. You said, "Now I know this is wrong." That is not true. Go ahead and substitute t = 0 and v = 2 into your last equation, and see what you get. Solve for C. Now continue solving the problem.