What is the distance covered by an object thrown in viscous matter?

[hastings]

an object with mass m=0.1kg is thrown with an initial velocity v0=20m/s in a viscous matter that exercises a resistant force of F=-Bv (B=2kg/s and v=velocity). ignoring the gravity force, find the distance covered by the object in the viscous medium.

I tried this
F=-Bv=ma => a=(-Bv)/m;
a=(dv)/dt => dv/dt=(-Bv)/m --> v dv=(-Bv)/m *dt
integrating I get

$$-\frac{B}{m}t=\log v - \log 20$$

since ds/dt=v

$$v=e^{-\frac{B}{m}t + \log 20}$$

then integrate again $$\int{ds}=\int {e^{-\frac{B}{m}t + \log 20}dt$$

 

[neutrino]

Hint: e^(a+b) = e^ae^b

 

[arunbg]

It would be much easier to use a = vdv/dx , than a=d²s/dt²

 

[hastings]

sorry didn't get you. dv/dx? How is it ? Am I on the right track?

$$s=\frac{C}{\alpha}e^{\alpha t}$$

where $$\alpha=-\frac{B}{m} \mbox{ and } C=e^{\log 20}$$

I know everything except t;

 

[arunbg]

No, i meant
$$a =v\times\frac{dv}{dx}$$

If you don't know how this equation arises, just try dividing its RHS numerator and denominator by dt.

 

[hastings]

Any hint that could help me solve this problem is appreciated.

 

[arunbg]

Okay let me make it a lot more simpler for you.
a=dv/dt
= (dx/dt)*dv/((dx/dt)*dt) {Multiplying numerator and denominator by dx/dt}
= v*dv/dx
Do u get me now?

 

[lalbatros]

Could be interresting to use the energy theorem.
The energy dissipated by the friction force is easy to calculate.

 

[Doc Al]

Note to the OP: Do Not post the same problem more than once!

https://www.physicsforums.com/showthread.php?t=171773