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Expert: Daniel Mazur - 2/10/2010


Question
Hi.
Can you help me with this somehow strange looking easy question by checking my answer:
a motion of an object is represented by this the position function of time
x(t)=a+b*e^ln(t)*e^-t
when does the object momentary stops:
v(t)=b(e^ln(t)/t e^-t + e^ln(t) e^-t *-1)
v(t)=0 ---->>> b(e^ln(t)/x e^-t - e^ln(t) e^-t )=0
(e^ln(t)/t e^-t - e^ln(l) e^-t )=0
1/t e^-t =e^ln(t) e^-t
1/t = e^ln(t)
1= e^ln(t)*t
1= e^ln(t) *ln(e^t)
e^ln(t)=ln(e^t)

first I am kind of stuck on math I can't remember the how to deal with logs second is this method correct I am can't focus about the physics about it since it seems all math question.

Answer
Hi Hamad,

you should use more parentheses, so the terms are absolutely clear. I interpret the given equation as x(t)=a+b*(e^ln(t))*(e^-t).
Before you start deriving, observe that there is an identity: e^ln(t)=t. You can substitute, just remember that the original equation with ln(t) imposed a limit on t, t>0.
Now x(t)=a+b*t*(e^-t), so the derivative v(t) = x(t)' = b*(e^-t)-b*t*(e^-t) = b*(e^-t)*(1-t). This equation has only one root v(t)=0 , which is at t=1.

I don't know, where your calculation went astray, but whether you use the identity to simplify the task, the result must be the same.

Take care!
Daniel

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