If a body moves through a liquid or a gas then the fluid applies a force on the body which is called drag force. Direction of the drag force is always opposite to the motion of the body relative to the fluid. At low speeds of the body, drag frog `( F _(P))` is directly proportional to the speed.
`F_(D) = kv `
What K is a proportionally constant and it depends upon the dimension of the body moving in air at relatively high speeds, the drag force applied by air an the body is proportional to `v^(2)`
Where this proportionally constant K can be given by
`K_(2) rho CA`
Where `rho` is the density of air
C is another constant givig the drag property of air
A is area of cross-section of the body
Consider a case an object of mass m is released from a height h and it falls under gravity. As it's speed increases the drag force starts increasing on the object. Due to this at some instant, the object attains equilibrium. The speed attained by the body at this instant is called "terminal speed" of the body.
Assume that the drag force applied by air on the body follows the relation `F_(D) = kv`,neglect the force by buoyancy applied by air on the body then answer the following questions.
What is the pattern of acceleration change of the body ?

Options

A. It first increases, then decreases to zero
B. It decreases uniformly from starting to zero
C. It decreases non-uniformly from starting to zero
D. It first increases then becomes constant

Answer

To analyze the pattern of acceleration change of a body falling through air under the influence of gravity and drag force, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces Acting on the Body**:
- When the body is released from height \( h \), it experiences two main forces:
- The gravitational force \( F_g = mg \) acting downwards.
- The drag force \( F_D = kv \) acting upwards, where \( v \) is the velocity of the body and \( k \) is a proportionality constant.

2. **Net Force and Acceleration**:
- The net force \( F_{net} \) acting on the body can be expressed as:
\[ F_{net} = F_g - F_D = mg - kv \]
- According to Newton's second law, the net force is also equal to mass times acceleration:
\[ F_{net} = ma \]
- Therefore, we can write:
\[ ma = mg - kv \]
- Rearranging gives us the expression for acceleration \( a \):
\[ a = g - \frac{k}{m}v \]

3. **Relating Acceleration to Velocity**:
- Since acceleration \( a \) can also be expressed as the time derivative of velocity:
\[ a = \frac{dv}{dt} \]
- We can substitute this into our previous equation:
\[ \frac{dv}{dt} = g - \frac{k}{m}v \]

4. **Separating Variables**:
- Rearranging gives us:
\[ \frac{dv}{g - \frac{k}{m}v} = dt \]
- This allows us to integrate both sides.

5. **Integrating**:
- We integrate from \( v = 0 \) to \( v \) and from \( t = 0 \) to \( t \):
\[ \int_0^v \frac{1}{g - \frac{k}{m}v} dv = \int_0^t dt \]
- The left side can be solved using the natural logarithm:
\[ -\frac{m}{k} \ln \left| g - \frac{k}{m}v \right| \Big|_0^v = t \]

6. **Finding Terminal Velocity**:
- As time approaches infinity, the velocity \( v \) approaches a terminal velocity \( v_t \) where the drag force equals the gravitational force:
\[ mg = kv_t \implies v_t = \frac{mg}{k} \]

7. **Behavior of Acceleration**:
- As \( v \) approaches \( v_t \), the term \( g - \frac{k}{m}v \) approaches zero, leading to:
\[ a \to 0 \]
- This indicates that the acceleration decreases over time and approaches zero as the body reaches terminal velocity.

### Conclusion:
The pattern of acceleration change of the body is that it starts at \( g \) (acceleration due to gravity) and decreases non-uniformly towards zero as the drag force increases with velocity. Eventually, the acceleration becomes zero when the drag force equals the weight of the body, resulting in terminal speed.