Acceleration in terms of position

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Homework Help Overview

The original poster is tasked with deriving an expression for acceleration as a function of time, given the acceleration equation a = -1.5*s, where s represents position. Initial conditions include a velocity of 20 m/s and a position of 0 m.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the Laplace transform and characteristic equations to derive expressions for position, velocity, and acceleration. They express concern about the relationship between acceleration and velocity, questioning the validity of their results when graphed.

Discussion Status

Some participants clarify that acceleration and velocity can indeed have opposite directions, suggesting that the original poster's confusion may stem from a misunderstanding of vector relationships. Others propose that the original poster should consider the implications of the equation a = -1.5*s for further insight.

Contextual Notes

The original poster is grappling with the implications of their derived functions, particularly regarding the behavior of velocity and acceleration in the context of their initial conditions and the nature of the equations involved.

yoamocuy
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Homework Statement


I'm given acceleration as a=-1.5*s where s is position, and I need to derive an expression for acceleration as a function of time. I am also given an initial velocity of 20 m/s and initial position of 0 m.


Homework Equations


a=-1.5*s

Characteristic Equation
s(t)=C1*e-p*t*cos(sqrt(q)*t)+C2e-p*t*sin(sqrt(q)*t)

The Attempt at a Solution


acceleration is the 2nd derivative of position, therefore a=-1.5*s is also equal to d2s/dt2=-1.5*s

d2s/dt2+1.5*s=0

I took the laplace transform of both sides to get: s2 + 1.5=0

Solving for s I get s=i*sqrt(1.5)

plugging this into the characteristic equation I get:
s(t)=C1*e0*cos(sqrt(1.5)*t)+C2*e0*sin(sqrt(1.5)*t)

at t=0 this equation becomes:
0=C1*e0*cos(0)+C2*e0*sin(0)

therefore C1=0

so s(t)=C2*sin(sqrt(1.5)*t)

Take the derivative to get:
v(t)=sqrt(1.5)*C2*cos(sqrt(1.5)*t)

at t=0 v(t)=20 therefore
20=sqrt(1.5)*C2*1

C2=16.33

that makes s(t)=16.33*sin(sqrt(1.5*t))
and v(t)=20*cos(sqrt(1.5)*t)

take the derivative to get a(t):
a(t)=-24.5*sin(sqrt(1.5)*t)

All of this seemed ok to me until I graphed all three functions and realized that according to these equations when my particle is accelerating its velocity is slowing down, which can not be possible. Did I do this question completely wrong?
 
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yoamocuy said:
All of this seemed ok to me until I graphed all three functions and realized that according to these equations when my particle is accelerating its velocity is slowing down, which can not be possible. Did I do this question completely wrong?
No. Acceleration and velocity are generally independent, and, in particular, they can have opposite directions. For example, if the initial velocity is vi in the positive x direction, and then the velocity changes by an amount Δv in the negative x direction, then the speed (magnitude of velocity) decreases. If this change happens in a time Δt, then the average acceleration is Δv/Δt in the negative x direction. Remember, velocity and acceleration are vectors, so they each have magnitude and direction.
 
I understand all that, but with my functions, since they are all cos and sin functions, when my particle's velocity is traveling in the negative direction my acceleration is traveling in the positive direction. It shouldn't be like that should it? Or wait.. I should be graphing this in radian rather than degrees shouldn't I?
 
It should be exactly like that, and you'll see why once you realize what the equation a=-1.5*s really means. If you use "x" instead of s and "F/m" instead of a, you'll get:

F/m=-1.5x
F=-1.5mx

Call 1.5m "k": F=-kx

Do you recognize that equation?
 
Oh ok, that makes it much clearer haha. Thanks
 

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