- #1
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?