# Homework Help: Acceleration in terms of velocity

1. Sep 7, 2009

### yoamocuy

Hey, so I have a question that states that a particle is subject to an acceleration of a=-3.0*v (m/s^2), and I need to find the acceleration, the velocity, and the position of the particle in terms of time. The initial velocity, Vo, is 15 m/s, and the initial position, So, is 0m.

Attempt at a Solution

I've tried plugging in a*t+Vo in for v, solving for a, and integrating, but that doesnt give me a velocity of 15 m/s at t=0s. Also I have tried solving for 3 equations 3 unknowns in terms of time using the equations a=-3.0*v, v=a*t+Vo, and d=1/2*a*t^2+Vo*t+So. This gave me the correct initial velocity and position at t=0; however, when I graphed all three functions it seemed that this could not be the correct method either because the particle was moving more meters in 1 second on my distance graph then its initial velocity at that second, which could not happen because it had a negative acceleration.

I'm thinking that I somehow need to get my acceleration in terms of time by setting v= to ds/dt and using a definite integral of some sort, but I just cant manage to figure out how to get it done. Any advice would be great.

2. Sep 7, 2009

### Nabeshin

Hello and welcome to PF!

You are onto something here. This problem is essentially one of differential equations, since the equation given to you (a=3v) is of that form. The first step is to write this as a differential equation for velocity, which you can solve to yield your velocity time function. Then you could either solve the second order differential equation from the a=3v, or integrate this velocity function, whichever you prefer.

3. Sep 8, 2009

### rl.bhat

You can solve by integration.
Now a = dv/dt = -3.0*v
Int.dv/v = -3.0*Int(dt)
log(v) = -3.0*t + logC At t = 0, log(v) = log(15) = log(C)
log(v/15) = -3.0*t Or
v =15 e^-3.0*t
Now v = dx/dt = 15*e^-3.0*t
Or dx= 15*e^-3.0*t*dt
Find the integration. To find the constant C, put x=0 for t =0.

4. Sep 8, 2009

### yoamocuy

thanks, that helps a lot