Acceleration in terms of velocity

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yoamocuy
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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 doesn't 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 can't manage to figure out how to get it done. Any advice would be great.
 
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yoamocuy said:
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 can't manage to figure out how to get it done. Any advice would be great.

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.
 
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.
 
thanks, that helps a lot