The Drag Force and Terminal Speed — Question 3

Question 3

A proposed speed curve must satisfy the force equation

Take $g=10$ m/s$^2$ and neglect buoyancy unless explicitly specified. A 1 kg body is released from rest in still fluid with linear drag $D=bv$, $b=2$ kg/s. Downward is positive. A proposed solution is $v(t)=5(1-e^{-2t})$ m/s with t in seconds.

1. Check the initial condition and the differential equation $m\,dv/dt=mg-bv$.

2. Find the time to reach 90 percent of terminal speed.

3. Integrate the speed to find the distance fallen by that time.

Question 3 - Solution

A proposed speed curve must satisfy the force equation

(a) Substitute rather than merely trust the curve.

The proposed expression satisfies the initial condition $v(0)=5(1-1)=0$. With time in seconds, $$\frac{dv}{dt}=10e^{-2t},\qquad g-\frac bm v=10-2[5(1-e^{-2t})]=10e^{-2t}.$$ Both sides agree, and $\boxed{v_t=mg/b=5\ \mathrm{m/s}}$.

(b) Use the remaining speed gap.

At 90 percent of terminal speed, $$.9=1-e^{-2t_{90}}\Longrightarrow e^{-2t_{90}}=.1.$$ $$\boxed{t_{90}=\frac{\ln10}{2}=1.1513\ \mathrm s}.$$

(c) Integrate with zero initial displacement.

Integrate from the release point: $$y(t)=\int_0^t5(1-e^{-2\tau})\,d\tau=5t-2.5(1-e^{-2t}).$$ $$\boxed{y(t_{90})=2.5\ln10-2.25=3.5065\ \mathrm m}.$$ Using $v_tt_{90}$ would omit the initial acceleration phase and overestimate the fall distance.