Acceleration, Velocity and distance

AI Thread Summary
An engineer is modeling a router's motion with an acceleration of -1.8 m/s², starting at a position of 0 m and a velocity of 2.4 m/s. The discussion revolves around integrating the acceleration to find the velocity and subsequently the position. The initial integration step must incorporate the initial velocity to determine the constant of integration correctly. The correct velocity function is derived as v = 2.4e^(-1.8t), which is then used to find the position function. The conversation highlights the importance of applying initial conditions in integration to achieve accurate 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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