Acceleration in terms of velocity

In summary, the particle experiences a=-3.0*v (m/s^2) acceleration and moves to a new position at t=0s according to the equation v=a*t+Vo.
  • #1
yoamocuy
41
0
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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  • #2
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.
 
  • #3
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
thanks, that helps a lot
 

What is acceleration?

Acceleration is the rate of change of velocity over time. It can be described as how quickly an object's velocity is changing, either by speeding up or slowing down.

How is acceleration related to velocity?

Acceleration and velocity are directly related. Increases in velocity result in positive acceleration, while decreases in velocity result in negative acceleration (also known as deceleration).

What are the units of acceleration?

The standard unit of acceleration is meters per second squared (m/s²). Other common units include kilometers per hour squared (km/h²) and centimeters per second squared (cm/s²).

How is acceleration measured?

Acceleration can be measured using a device called an accelerometer. This device measures changes in velocity and calculates the acceleration based on the time it takes for the change to occur.

What are some examples of acceleration in everyday life?

Examples of acceleration in everyday life include a car speeding up on the highway, a person jumping off a diving board, and a ball rolling down a hill. Any time an object's velocity changes, there is acceleration involved.

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