Which of the following option correctly describes the variation of the speed ` ` and acceleration `'a'` of a point mass falling vertically in a viscous medium that applies a force `F=-kv`, where `'k'` is constant, on the body?
(Graphs are schematic and drawn to scale)

Options

A.
B.
C.
D.

Explanation

To solve the problem of determining the variation of speed \( v \) and acceleration \( a \) of a point mass falling vertically in a viscous medium, we can follow these steps:

### Step 1: Understand the Forces Acting on the Mass
The forces acting on the mass are:
- The gravitational force \( F_g = mg \) acting downwards.
- The viscous drag force \( F_d = -kv \) acting upwards.

### Step 2: Write the Equation of Motion
Using Newton's second law, we can set up the equation of motion:
\[ ma = mg - kv \]
Rearranging gives:
\[ ma = mg - kv \implies a = g - \frac{k}{m}v \]

### Step 3: Identify Terminal Velocity
At terminal velocity \( v_t \), the acceleration \( a \) becomes zero:
\[ 0 = g - \frac{k}{m}v_t \implies v_t = \frac{mg}{k} \]

### Step 4: Express Acceleration in Terms of Velocity
From the equation \( a = g - \frac{k}{m}v \), we can see that as \( v \) approaches \( v_t \), the acceleration \( a \) approaches zero.

### Step 5: Analyze the Behavior of Velocity Over Time
To find how velocity changes with time, we can rearrange the equation:
\[ \frac{dv}{dt} = g - \frac{k}{m}v \]
This is a first-order linear differential equation. Solving this will show that velocity increases towards \( v_t \) exponentially.

### Step 6: Solve the Differential Equation
Separating variables and integrating:
\[ \int \frac{1}{g - \frac{k}{m}v} dv = \int dt \]
This leads to:
\[ t = -\frac{m}{k} \ln\left( \frac{g - \frac{k}{m}v}{g} \right) \]
From this, we can derive that:
\[ v(t) = v_t \left(1 - e^{-\frac{k}{m}t}\right) \]
This shows that velocity increases exponentially and approaches \( v_t \).

### Step 7: Determine the Behavior of Acceleration
Since \( a = g - \frac{k}{m}v \) and \( v \) increases over time, \( a \) decreases over time and approaches zero as \( v \) approaches \( v_t \).

### Final Conclusion
- The velocity \( v \) increases exponentially and approaches \( v_t \).
- The acceleration \( a \) decreases exponentially and approaches zero.