Problem 3.58 Challenge

พิจารณา F.B.D. ของวัตถุมวล m (ให้ทิศลงเป็น \(+\) เพื่อความสะดวก)

\(R = bv\) จาก \(ΣF_{y}= ma_{y}\)

\(vmg - bv = ma\)

a

\(mgmg - bv = m\frac{dv}{dt}\)

\(\frac{dv}{g -\frac{bv}{m}}= dt\)

\(_{vt}\)

\(\int \frac{dv}{g -\frac{bv}{m}}=\int dt\)

\(_{00}\)

กำหนดให้ \(u = g - \frac{bv}{m}\)

\(_{vt}\)

จะได้ \(-\frac{m}{b}\ln (g -\frac{bv}{m})=t\)

\(_{00}\)

\(\ln (1 -\frac{bv}{mg}) = -\frac{bt}{m}\)

\(_{-bt/m}\)

\(v =\frac{mg}{b}(1 - e)\)

v

จะได้

\(\frac{_{mg}}{_{b}}\)

\(_{0.632}\frac{_{mg}}{_{b}}\)

\(t\;\blacksquare\)

\(\frac{_{m}}{_{b}}\)