Classical-Mechanics · Unit 10 · Video 3 · Interactive Practice

The Reciprocal Tail: Velocity Decay Under High-Speed Quadratic Drag

IKey Formulas

Formula	Name	What you need
$m\dfrac{dv}{dt} = -\beta v^2$	Equation of motion (quadratic drag)	Mass $m$ , drag coefficient $\beta$
$\beta = \tfrac{1}{2} C_D A \rho$	Drag coefficient	Shape $C_D$ , area $A$ , density $\rho$
$v(t) = \dfrac{v_0}{1 + t/\tau}$	Velocity (reciprocal decay)	Launch speed $v_0$ , time constant $\tau$
$\tau = \dfrac{m}{\beta v_0}$	Time constant	$m$ , $\beta$ , $v_0$

Key Insight: The force law dictates the decay law — squaring the velocity in the drag turns exponential decay into a slower, lingering $1/t$ reciprocal tail.

IIThe Time Constant Sets the Timescale

The time constant $\tau = m/(\beta v_0)$ carries the mass — how does a heavier object reshape the decay?

Mass m (kg) 2

IIITwo Fingerprints: Reciprocal vs Exponential

Early on the two decay laws look identical — where do their tails part ways?

💡 For $t \gg \tau$ the reciprocal law approaches $v \approx \dfrac{v_0\,\tau}{t}$ — a $1/t$ power-law tail — while the exponential $v_0 e^{-t/\tau}$ is already vanishingly small.

IVPredict the Speed Deep in the Tail

Before computing: how much speed survives at $t = 3\tau = 6$ s, out in the tail?

Your guess: v(6 s) 10.0

VQuiz Questions

Problem 1 · Evaluate the Solution

Given: quadratic drag gives $v(t) = \dfrac{v_0}{1 + t/\tau}$ with $v_0 = 20$ m/s and $\tau = 2$ s — find $v$ at $t = 2$ s.

✅ Correct! At

t = \tau

the denominator is

1 + 1 = 2

, so the speed is exactly half of

v_0

❌ Wrong decay law.

7.4

m/s is

v_0 e^{-t/\tau}

— that is the exponential (linear-drag) answer. Quadratic drag decays as

1/(1+t/\tau)

❌ Not quite. Substitute carefully:

1 + t/\tau = 1 + 2/2 = 2

, not

1 + t

Show solution

Substitute $t = 2$ s, $\tau = 2$ s into the reciprocal solution:

v(2) = \frac{v_0}{1 + t/\tau} = \frac{20}{1 + \tfrac{2}{2}} = \frac{20}{2} = 10\ \text{m/s}

At exactly one time constant the speed has dropped to $v_0/2$ . (An exponential would give $20\,e^{-1} \approx 7.4$ m/s — it plunges faster.)

Problem 2 · Does Mass Cancel?

Given: two projectiles fired with the same $v_0$ into the same fluid (same $\beta$ ). Projectile A has twice the mass of projectile B — which slows more gradually?

✅ Correct!

\tau \propto m

, so

\tau_A = 2\tau_B

; a larger time constant stretches

v(t) = v_0/(1+t/\tau)

into a gentler decay.

❌ Mass does not cancel. Dividing

m\,\dfrac{dv}{dt} = -\beta v^2

m

gives

\dfrac{dv}{dt} = -\dfrac{\beta}{m}v^2

— the mass sets the deceleration and survives inside

\tau = m/(\beta v_0)

❌ Not quite. Compare the time constants:

\tau = m/(\beta v_0)

grows with mass, and a larger

\tau

means slower decay.

Show solution

The time constant is $\tau = \dfrac{m}{\beta v_0}$ . With $v_0$ and $\beta$ fixed, $\tau$ is proportional to mass:

\frac{\tau_A}{\tau_B} = \frac{m_A}{m_B} = 2

Since $v(t) = v_0/(1 + t/\tau)$ decays more slowly for larger $\tau$ , the heavier projectile A slows more gradually. Physically, both feel the same drag force at a given speed, but A's larger inertia resists the deceleration.

Problem 3 · From Constants to a Time

Given: $m = 4$ kg, $\beta = 0.05$ kg/m, $v_0 = 20$ m/s, decaying as $v(t) = v_0/(1 + t/\tau)$ — find $\tau$ , then the time when $v = 8$ m/s.

What is the time constant $\tau$ ?

At what time does $v = 8$ m/s?

✅ Correct!

\tau = 4

s, and inverting

v(t) = 8

gives

t = \tau\!\left(\tfrac{v_0}{v} - 1\right) = 4(2.5 - 1) = 6

❌ Check the time constant.

\tau = \dfrac{m}{\beta v_0}

— keep all three factors: dividing only by

\beta

gives

80

s (you dropped

v_0

); inverting the ratio gives

0.25

❌ Check the time. Solve

\dfrac{20}{1 + t/\tau} = 8 \Rightarrow 1 + t/\tau = 2.5

. Don't drop the

\tau

(

t = \tau \cdot 1.5

), and don't forget the

-1

Show solution

Step 1 — time constant:

\tau = \frac{m}{\beta v_0} = \frac{4}{(0.05)(20)} = \frac{4}{1} = 4\ \text{s}

Step 2 — solve for the time at $v = 8$ m/s:

\frac{20}{1 + t/\tau} = 8 \;\Rightarrow\; 1 + \frac{t}{4} = \frac{20}{8} = 2.5 \;\Rightarrow\; \frac{t}{4} = 1.5 \;\Rightarrow\; t = 6\ \text{s}

(An exponential would reach $8$ m/s at $t = \tau\ln 2.5 \approx 3.7$ s — sooner, because it decays faster.)

Problem 4 · The Shape of the Tail

Given: $v(t) = \dfrac{v_0}{1 + t/\tau}$ — at late times $t \gg \tau$ , how does the speed fall off?

✅ Correct! For

t \gg \tau

the

1

is negligible, so

v \approx v_0\tau/t

— a

1/t

power law, the long reciprocal tail.

❌ That is the other law.

e^{-t/\tau}

is the linear-drag fingerprint. Quadratic drag never gives an exponential — its tail is the much slower

1/t

❌ Not quite. Drop the

1

1 + t/\tau

for large

t

v \approx v_0\tau/t

, which decays as

1/t

(not

1/t^2

, and it does reach zero).

Show solution

For $t \gg \tau$ the constant term is dwarfed by $t/\tau$ :

v(t) = \frac{v_0}{1 + t/\tau} \;\xrightarrow{\,t \gg \tau\,}\; \frac{v_0}{t/\tau} = \frac{v_0\,\tau}{t} \;\propto\; \frac{1}{t}

The velocity decays as a $1/t$ power law — approaching zero but far more slowly than the exponential $e^{-t/\tau}$ of low-speed linear drag. This lingering $1/t$ behavior is the "reciprocal tail."

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