An object is being dropped from a height \(h\) above the ground. Apart from the force of gravity, an additional drag force \(F=-kv\) acts on the object. Find the correct graph of velocity \(v\) versus time \(t\).
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Start from the equation of motion $m\dfrac{dv}{dt} = mg - kv$ and notice that the net force, and so the acceleration, keeps shrinking as v grows, which rules out any graph showing a straight line. Velocity should rise quickly at first and then bend and flatten out, approaching a fixed terminal velocity without ever reaching it in a sudden jump. Use this curving, flattening shape to eliminate straight line graphs and any graph that reaches a constant value abruptly.
Concept:
When a body falls under gravity with a resistive force proportional to velocity:
Gravitational force \(= mg\) acts downward.
Drag force \(= kv\) acts upward (opposite to motion).
Net force decreases as velocity increases.
The equation of motion is:
\[
m\frac{dv}{dt} = mg - kv
\]
Step 1: Analyze the differential equation.
\[
\frac{dv}{dt} = g - \frac{k}{m}v
\]
This is a first-order linear differential equation whose solution is:
\[
v(t) = \frac{mg}{k}\left(1 - e^{-\frac{k}{m}t}\right)
\]
Step 2: Study the nature of the velocity–time graph.
From the solution:
At \(t=0\), \(v=0\) (object is dropped from rest).
Velocity increases with time.
The slope \(\dfrac{dv}{dt}\) decreases continuously.
Velocity approaches a constant value called terminal velocity:
\[
v_t = \frac{mg}{k}
\]
Step 3: Match with the given graphs.
The correct graph must:
Start from the origin.
Increase monotonically.
Gradually flatten and approach a horizontal asymptote.
This behavior corresponds to Graph 2.
Conclusion:
\[
\boxed{\text{Graph 2 correctly represents } v \text{ versus } t}
\]
Hence, the correct answer is (B).
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Approach Solution -2
Concept:
There is no need to solve the differential equation to identify the graph. Just track how the acceleration changes as velocity builds up.
The second law of motion directly tells us the slope of the $v$-$t$ graph at every instant, without integrating anything.
Step 1: Write the net force at any instant.
$ma = mg - kv$, so $a = g - \dfrac{k}{m}v$.
Step 2: Check the slope at $t = 0$.
At the start, $v = 0$, so $a = g$. The graph must begin with its steepest slope, not a flat or curved start.
Step 3: Track how the slope changes as $v$ increases.
As $v$ grows, $kv$ grows, so $a = g - \dfrac{k}{m}v$ keeps decreasing. This means the slope of the $v$-$t$ curve keeps falling — the curve can never be a straight line or bend upward.
Step 4: Find where the slope becomes zero.
$a=0$ when $v = v_t = \dfrac{mg}{k}$ (terminal velocity). Beyond this point the graph flattens into a horizontal line.
Step 5: Match the shape.
A curve that starts steep, keeps bending downward, and flattens out smoothly without overshooting matches Graph 2.
