Homework Help: Acceleration in terms of position

1. Sep 8, 2009

yoamocuy

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
a=-1.5*s

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

3. 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?

2. Sep 8, 2009

turin

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.

3. Sep 8, 2009

yoamocuy

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?

4. Sep 8, 2009

ideasrule

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?

5. Sep 8, 2009

yoamocuy

Oh ok, that makes it much clearer haha. Thanks