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 \(t\).
Show Hint
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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Show Solution
Approach Solution -2
Concept:
Instead of tracking the slope continuously, just look at the two extreme situations: right after release, and after a long time. Any curve consistent with both extremes, and with no sudden jumps in between, has to be the answer.
Step 1: Examine the situation just after release.
Velocity is nearly zero, so drag $kv$ is negligible. The object falls almost freely, so $v \approx gt$ — a steep, nearly straight rise at the start.
Step 2: Examine the situation after a long time.
As $v$ grows, drag grows too, until $kv = mg$. At this point the net force is zero, so $v$ stops changing and settles at the terminal value $v_t = \dfrac{mg}{k}$.
Step 3: Connect the two extremes.
This is a simple first-order equation with only one power of $v$ in the drag term, so there is no oscillation or overshoot between these two extremes: the curve rises smoothly from the steep start to the flat finish.
Step 4: Eliminate the wrong options.
A straight line would mean constant acceleration (ignores drag entirely) — wrong.
A curve that keeps rising without flattening ignores the terminal velocity limit — wrong.
A curve that overshoots and comes back down does not match a monotonically increasing velocity — wrong.
Only a curve that starts steep and bends smoothly into a flat line fits both extremes.
