Question

A particle of mass \(m\) falls from rest through a resistive medium having resistive force \(F=-kv\), where \(v\) is the velocity of the particle and \(k\) is a constant. Which of the following graphs represents velocity \(v\) versus time \(t\)? 

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

Hint:

Whenever resistive force is proportional to velocity, the speed approaches terminal velocity exponentially, not linearly.

Answer 1

The Correct Option is A

Concept: When a body falls through a resistive medium with resistive force proportional to velocity, the motion is governed by a first-order linear differential equation. The velocity increases with time but approaches a constant maximum value called {terminal velocity}.
Step 1: Write the equation of motion For a particle falling downward, forces acting are:
Weight \(mg\) (downward)
Resistive force \(kv\) (upward, opposing motion) Taking downward direction as positive: \[ m\frac{dv}{dt} = mg - kv \]
Step 2: Rearrange the equation \[ \frac{dv}{dt} + \frac{k}{m}v = g \] This is a first-order linear differential equation.
Step 3: Solve the differential equation Using integrating factor: \[ \text{I.F.} = e^{\frac{k}{m}t} \] Solution: \[ v(t)=\frac{mg}{k}\left(1-e^{-\frac{k}{m}t}\right) \]
Step 4: Analyze the nature of the solution From the expression:
At \(t=0\): \[ v(0)=0 \quad \text{(starts from rest)} \]
As \(t\to\infty\): \[ v\to \frac{mg}{k} \quad \text{(terminal velocity)} \] Thus:
Velocity increases rapidly at first
Rate of increase gradually decreases
Velocity approaches a constant value asymptotically
Step 5: Match with the given graphs The correct \(v\)–\(t\) graph must:
Start from the origin \((0,0)\)
Rise monotonically
Approach a horizontal asymptote (terminal velocity) Among the given options:
Graph 1 shows velocity increasing and gradually flattening out
Graph 2 shows increasing curvature without saturation
Graph 3 shows linear increase (no resistance)
Graph 4 shows decreasing velocity Hence, Graph 1 correctly represents the motion.