A small, spherical bead of mass 3.00 g is released from rest at t 5 0 from a poi

Question

A small, spherical bead of mass 3.00 g is released from rest at t 5 0 from a point under the surface of a viscous liq-uid. The terminal speed is observed to be v T 5 2.00 cm/s. Find (a) the value of the constant b that appears in Equa-tion 6.2, (b) the time t at which the bead reaches 0.632v T , and (c) the value of the resistive force when the bead reaches terminal speed.

I just need help with part b (differential equations).
I understand up to this point..

[(m)/(mg-bv)]dv = dt

I dont understand this part of the solution..
mln(mg-bv) = -bt + C

Where did the -b come from in -bt? Please answer that and explain this math the best that you can. thanks

Explanation / Answer

[(m)/(mg-bv)]dv = dt dv/dt=g-bv/m Put - bv/m=V dv=-dV*m/b -dV*m/b=gdt+Vdt dV*m=-bgdt-bVdt dV*m/(g+V)=-bdt itigrate mln(g+V)=-bt+C put V=-bv/m again mln(mg-bv) = -bt + C 1 [c1=c+mln(m)] hope now u got it....

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.