Acceleration, Velocity and distance

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Homework Help Overview

The discussion revolves around a problem in kinematics involving the acceleration, velocity, and position of a router in a machining process. The original poster presents a scenario where the router's acceleration is given, along with initial conditions for its position and velocity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of acceleration to find velocity and subsequently position. There is a focus on the correct application of initial conditions and constants of integration. Questions arise regarding the validity of the steps taken and the use of initial velocity in the integration process.

Discussion Status

Participants are actively engaging with each other's attempts, providing feedback on integration steps and the use of initial conditions. There is a recognition of the need to incorporate the initial velocity into the calculations, and some guidance has been offered regarding the integration process.

Contextual Notes

There is an ongoing discussion about the correct application of initial conditions and the integration constants, with some participants expressing uncertainty about their methods and results.

Oblivion77
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Homework Statement



An engineer designing a system to control a router for a machining process models the system so that the router's acceleration during an interval of time is a=-1.8vm/s . When t=0 , its position is s=0 and its velocity is 2.4 m/s. Determine the router's position (unit: m) at s.

Homework Equations



a=dv/dt, v=ds/dt

The Attempt at a Solution



First i did -1.8v=dv/dt did some integration and work and found v = e^(-1.8t)

I then used e^(-1.8t) = ds/dt and solved for s and subsituted t=0(lower bound) and t=0.75(upper bound) and found s = 0.412m

I am not sure if the first step I did is allowed since I never used the v=2.4. Please tell me my mistakes. Thanks!
 
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Oblivion77 said:
First i did -1.8v=dv/dt did some integration and work and found v = e^(-1.8t)

I then used e^(-1.8t) = ds/dt and solved for s and subsituted t=0(lower bound) and t=0.75(upper bound) and found s = 0.412m

I am not sure if the first step I did is allowed since I never used the v=2.4. Please tell me my mistakes. Thanks!

Hi Oblivion77! :smile:

You integrated twice, but you only put in a constant of integration the second time.

The v = 2.4 comes in the first time! :wink:
 
Ok, using the v=2.4 with the first integration I get -1.8v=dv/dt, doing some integration I get
t=(-1/1.8)ln(v)+0.49 (does this look right so far?). Now I want to solve for v and use v=ds/dt to get the function for position. so v=e^(0.88-1.8t), e^(0.88-1.8t)=ds/dt .Then to sub t=0.75s for the upper bound and t=0 for the lower bound. Does this look right now? Or did I make a mistake somewhere? Thanks!
 
Oblivion77 said:
Ok, using the v=2.4 with the first integration I get -1.8v=dv/dt, doing some integration I get
t=(-1/1.8)ln(v)+0.49

oooh :cry:

No, you integrated dv/v - -1.8dt to get v = Ce-1.8t.

And so, when t= 0, e-1.8t = … ? , and so C = … ? :smile:
 
Ok, so when i do that I get v=Ce^(-1.8t). And the intial conditions are t=0, v=2.4. So would i do 2.4=Ce^0, C=2.4. Then v=2.4e^(-1.8t), or do i not use the velocity?
 
What a long question! :wink:

Yes, v = 2.4e-1.8t

why does that worry you? :smile:
 

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