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
For motion with linear drag:
Velocity never increases linearly.
Terminal velocity is always reached asymptotically.
Curvature in \(v\)-\(t\) graph indicates decreasing acceleration.
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).
