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Worked solution

Keep both sides of a terminal velocity when separating

[Motion of Particles with Resistance]

Modified from Fundamentals of Differential Equations and Boundary Value Problems, sixth edition, p110. An object starts with velocity v0 and falls under gravity g=9.81g=9.81 metres per second squared. Air resistance gives deceleration proportional to velocity. Given dv/dt=g−kvdv/dt=g-kv, k>0k>0, express v in terms of t.

CJC · Lecture · Example 12

Separate and integrate the non-equilibrium case

∫dv/(9.81−kv)=∫1 dt\int dv/(9.81-kv)=\int1\,dt gives −ln⁡∣9.81−kv∣/k=t+C-\ln|9.81-kv|/k=t+C. Hence ln⁡∣9.81−kv∣=−kt−kC\ln|9.81-kv|=-kt-kC and 9.81−kv=Be−kt9.81-kv=Be^{-kt}, absorbing either sign into B.

How did this part go?

Fit the initial velocity and restore the equilibrium

At t=0, B=9.81−kv0B=9.81-kv_0. Therefore v=9.81/k−(9.81/k−v0)e−ktv=9.81/k-(9.81/k-v_0)e^{-kt}. This also includes the equilibrium v0=9.81/kv_0=9.81/k, where separation by 9.81−kv9.81-kv was not valid. The source remark says the initial side of the terminal value is not known from the question, so the sign must be retained. No assumption that v0 lies below terminal velocity is imposed; all these solutions approach 9.81/k.

How did this part go?

Common mistakes

Dividing by g−kv can lose the constant terminal-velocity solution.

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