An object falling through a fluid is observed to have acceleration given by $a = g - bv$ where $g$ = gravitational acceleration and $b$ is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?

An object falling through a fluid is observed to h – Class 11 Physics

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a = g - bv

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Concept understanding — Instantaneous Velocity

Instantaneous Velocity: From "How Fast" to "How Fast Right Now"

You already know average velocity. If a car travels 120 km in 2 hours, its average velocity is 60 km/h. That tells you the overall rate, but it hides everything that happened in between — the traffic jams, the sudden bursts of speed, the moments the car was completely stopped.

Now imagine you want to know the car's velocity at exactly 10:15 AM, not averaged over an hour or a minute. That's instantaneous velocity — the velocity at a single instant of time.

The Intuition: Zooming In

Think of a speedometer needle. When you drive, the needle doesn't stay fixed at 60 km/h. It jumps up when you accelerate, drops when you brake. At any given moment, the needle points to a specific number. That number is your instantaneous speed (velocity, if direction matters).

But here's the puzzle: at a single instant, the car hasn't moved any distance. How can you have a speed if $\Delta t = 0$ ? You can't divide by zero.

The trick is to shrink the time interval smaller and smaller, and see what the average velocity approaches.

The Precise Definition

Let $s(t)$ be the position of an object at time $t$ . The average velocity over a time interval $[t, t+h]$ is:

v_{\text{avg}} = \frac{s(t+h) - s(t)}{h}

Now, let $h$ get closer and closer to 0 (but never equal to 0). If the average velocity settles down to a single number as $h \to 0$ , that number is the instantaneous velocity at time $t$ :

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

This limit is exactly the derivative of position with respect to time. In calculus notation: $v(t) = s'(t)$ .

A Concrete Example

Suppose a ball is dropped from rest, and its height (in meters) after $t$ seconds is $s(t) = 4.9t^2$ (ignoring air resistance).

Average velocity from $t=2$ to $t=2.1$ seconds:

v_{\text{avg}} = \frac{4.9(2.1)^2 - 4.9(2)^2}{0.1} = \frac{4.9(4.41 - 4)}{0.1} = \frac{4.9 \times 0.41}{0.1} = 20.09 \text{ m/s}

Average velocity from $t=2$ to $t=2.01$ :

v_{\text{avg}} = \frac{4.9(2.01)^2 - 4.9(2)^2}{0.01} = \frac{4.9(4.0401 - 4)}{0.01} = 19.649 \text{ m/s}

Average velocity from $t=2$ to $t=2.001$ :

v_{\text{avg}} = \frac{4.9(2.001)^2 - 4.9(2)^2}{0.001} = 19.6049 \text{ m/s}

The numbers are converging to 19.6 m/s. That's the instantaneous velocity at $t=2$ seconds.

Using the derivative: $v(t) = 9.8t$ , so $v(2) = 19.6$ m/s. Matches perfectly.

Key Takeaways for Exams

Concept	Meaning	Formula
Average velocity	Total displacement ÷ total time	$\frac{\Delta s}{\Delta t}$
Instantaneous velocity	Velocity at a single moment	$\lim_{h \to 0} \frac{s(t+h)-s(t)}{h}$

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The constant speed is the terminal velocity, found by setting acceleration to zero in $a = g - bv$ , giving $v_t = g/b$ .

The key idea here is that "constant speed" means zero acceleration. When an object falls through a fluid, two forces act: gravity pulling it down, and a drag force opposing motion. The given equation $a = g - bv$ already encodes this physics — it's Newton's second law per unit mass. The term $g$ is the acceleration due to gravity, and $-bv$ is the deceleration caused by drag, which grows with speed.

At the moment of release, speed is zero, so drag is zero and acceleration is $g$ . As speed increases, drag grows, reducing the net acceleration. Eventually, when drag exactly balances gravity, the net force becomes zero and acceleration drops to zero. The object then falls at a constant speed — this is terminal velocity.

Set acceleration to zero because constant speed means no change in velocity:

a = 0

Substitute into the given equation:

0 = g - bv

Solve for $v$ :

bv = g \quad \Rightarrow \quad v = \frac{g}{b}

That's it — the constant speed is simply $g/b$ . …

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Concept: Terminal Velocity as the Asymptote of the Full Solved Motion, Not a Shortcut Substitution

Method: Solve the Differential Equation $\dfrac{dv}{dt}=g-bv$ Exactly, Then Take the Long-Time Limit

The stored answer takes a direct shortcut — set $a=0$ and solve algebraically for $v$ . This method instead does the harder, more rigorous thing the question is literally asking for ("after a long time... what speed"): it solves the full differential equation for $v(t)$ and takes $t\to\infty$ , which both confirms the shortcut's answer and reveals that the object only approaches this speed, never exactly reaching it in finite time.

Steps

Set up the separable differential equation from the given acceleration:

\frac{dv}{dt} = g - bv \quad\Longrightarrow\quad \frac{dv}{g-bv} = dt

Integrate both sides. The left side needs a substitution $w=g-bv$ , $dw=-b\,dv$ :

\int \frac{dv}{g-bv} = -\frac{1}{b}\ln|g-bv| + C_1

So:

-\frac{1}{b}\ln|g-bv| = t + C

Apply the initial condition — released from rest, $v(0)=0$ :

-\frac{1}{b}\ln|g| = C \quad\Longrightarrow\quad -\frac{1}{b}\ln|g-bv| = t - \frac{1}{b}\ln g

Solve for $v(t)$ explicitly. Rearranging:

\ln\left(\frac{g-bv}{g}\right) = -bt \quad\Longrightarrow\quad g-bv = g\,e^{-bt} \quad\Longrightarrow\quad v(t) = \frac{g}{b}\left(1-e^{-bt}\right)

(Note the structural resemblance to $x_0(1-e^{-\gamma t})$ elsewhere in this chapter — this is the same exponential-relaxation pattern, here governing speed approaching a limiting value instead of position approaching a limiting displacement.)

Take the long-time limit, exactly as the question asks ("after a long time"). As $t\to\infty$ , $e^{-bt}\to0$ : $\lim_{t\to\infty} v(t) = \frac{g}{b}(1-0) = \frac{g}{b}$ …

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Q.An object falling through a fluid is observed to have acceleration given by a=g−bva = g - bva=g−bv where ggg = gravitational acceleration and bbb is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?

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Q.An object falling through a fluid is observed to have acceleration given by $a = g - bv$ where $g$ = gravitational acceleration and $b$ is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?

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