Question

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 \(t\).

• A. Graph 1
• B. Graph 2
• C. Graph 3
• D. Graph 4

Hint:

For motion with linear drag:

Velocity never increases linearly.
Terminal velocity is always reached asymptotically.
Curvature in \(v\)-\(t\) graph indicates decreasing acceleration.

Answer 1

The Correct Option is B

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).