Acceleration in terms of position

AI Thread Summary
The discussion revolves around deriving an expression for acceleration as a function of time given the equation a = -1.5*s, where s is the position. The user successfully applies the Laplace transform and finds the position and velocity equations, but is confused by the relationship between acceleration and velocity, particularly when they appear to be in opposite directions. It is clarified that acceleration and velocity are independent vectors and can indeed act in opposite directions, which is consistent with the derived equations. The analogy to Hooke's Law (F = -kx) helps to clarify the physical interpretation of the acceleration equation. The user gains a better understanding of the dynamics involved in the motion described by the equations.
yoamocuy
Messages
37
Reaction score
0

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?
 
Physics news on Phys.org
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top