Motion with frictional force but without driving force

[kuruman]

should i bother about v₀?

Not yet. Do you recognize the ratio in the argument of the log for what it is? Specifically, can you simplify the numerator some?

 

[LockeZz]

i think i can convert to tangent:

$$x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))$$

then simplify again:

$$x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))$$

 

[kuruman]

Note that if you let v₀ become very large, you get an infinity. I hadn't appreciated that in the beginning. One thing that bothers me is the negative sign up front. Certainly x must be positive and the argument of the log is greater than 1 especially if v₀ becomes very large. Check your integral tables and don't forget that you are integrating

$$\int Tan(\delta - t)dt$$

There is a negative sign in front of t.

Check that and you are done.

 

[LockeZz]

hm.. does that mean that the negative sign represent a mistake in the expression?

 

[kuruman]

All I am saying is that

$$<br /> \int Tan(\delta - t)dt=Log[Cos(t-\delta)]<br />$$

 

[LockeZz]

I shall check back from the beginning the see if any mistake again.. thanks for the guidance by the way.. i really appreciate on your help.. Thank you very much :)