# Finding Uncertainty of the Second Derivative

Hey there,

I have a problem to do, in which I need to determine the acceleration of an object and the uncertainty of the acceleration. The position vs. time equation is given by:

s(t) = 0.205t2 + 0.3001t

Therefore, after differentiation, I can state the the velocity vs. time equation is:

v(t) = 0.410t + 0.3001

And that the acceleration of my object is:

a(t) = 0.410

The acceleration is constant, and this is the number that I'm looking for. However, I don't know how to calculate the uncertainty of the acceleration.

I know that the uncertainty of the position for the s(t) equation is +/- 0.005, and that the uncertainty of the time is +/- 0.15. How do I determine the uncertainty of the acceleration?

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The way I understand propagation of uncertainty, you'll use the following analysis:

Position function: s = s0 + v0*t + (a*t^2)/2
It looks like s0 is zero, so we get: s = v0*t + (a*t^2)/2
The uncertainty can be calculated as: u(s) = u[v0*t+(at^2)/2]
=> u(v0*t) = v0*t*SQRT[(u(v0)/v0)^2+(u(t)/t)^2]
=> substitute into: u(s) = u(v0*t)+u(at^2/2)
=> u(at^2/2)=1/2 u(a*t^2)
=> u(at^2) = a*t^2*SQRT[(u(a)/a)^2+(u(t^2)/t^2)^2]
=> u(t^2) = 2*t*u(t)

This makes the total uncertainty:

u(s) = v0*t*SQRT[(u(v0)/v0)^2+(u(t)/t)^2]+1/2*a*t^2*SQRT[(u(a)/a)^2+(2*u(t)/t)^2]

You can solve for the u(a) from this equation, but you might need the u(v0). If you don't have it, then I think you can just use u(v0)=0. That would be my approach. I used the equations from the following website: http://www.physics.uc.edu/~bortner/...pendix 2/Appendix 2 Error Propagation htm.htm

When you write v0, do you mean v-not, a.k.a. v0?