The Drag Force and Terminal Speed — Question 1

Question 1

Terminal speed is a balance, not a speed limit

Take $g=10$ m/s$^2$ and neglect buoyancy unless explicitly specified. A 2 kg body moves vertically in still fluid. Drag has magnitude $D=cv^2$, opposite velocity, with $c=0.20$ kg/m. Its motion is downward in all three cases considered.

1. Find terminal speed.

2. Find signed downward acceleration at 5, 10 and 15 m/s.

3. Explain whether a body already moving at 15 m/s violates the drag model and how it approaches terminal speed.

Question 1 - Solution

Terminal speed is a balance, not a speed limit

(a) Set the net force to zero.

Take downward positive. The force equation is $$ma=mg-cv^2.$$ At terminal motion, $a=0$, so $$\boxed{v_t=\sqrt{\frac{mg}{c}}=\sqrt{\frac{2(10)}{.20}}=10\ \mathrm{m/s}}.$$

(b) Keep the downward sign convention.

For downward motion, evaluate $$a(v)=g-\frac cmv^2=10-.10v^2\quad\text{in SI numerical units}.$$ $$\boxed{a(5)=+7.5,\qquad a(10)=0,\qquad a(15)=-12.5}\quad\mathrm{m/s^2}.$$ Negative acceleration at 15 m/s means upward acceleration while the body is still descending.

(c) Distinguish equilibrium from a bound.

For $v\geq0$, the equation can be factored as $$\frac{dv}{dt}=\frac cm(v_t-v)(v_t+v).$$ Thus $v>v_t$ gives $dv/dt<0$, whereas $v<v_t$ gives $dv/dt>0$. Terminal speed is a stable balance, not an upper bound on initial speed. This smooth model approaches it asymptotically unless it starts exactly there.