This problem involves a body falling under the influence of gravity with air resistance, which is proportional to the square of the velocity. The drag force is given as \( D = bv^2 \), where \( b \) is a constant, and \( v \) is the velocity of the body. Let’s go through each part of the problem step by step to solve for the required quantities.
Given Data
- The mass of the body is \( m \),
- The drag force is \( D = bv^2 \),
- The acceleration due to gravity is \( g \).
Part (a): Initial Downward Acceleration
At the initial moment, the body is at rest, meaning \( v = 0 \). Therefore, there is no drag force acting on the body at \( t = 0 \), and the only force acting on the body is the gravitational force, which causes the downward acceleration.
The gravitational force is:
F_{\text{gravity}} = mg
The net force on the body at the start is equal to the gravitational force since the drag force is zero at this point. According to Newton’s second law, the acceleration is:
a_{\text{initial}} = \frac{F_{\text{net}}}{m} = \frac{mg}{m} = g
Thus, the initial downward acceleration of the body is:
a_{\text{initial}} = g
So, the body initially falls with the acceleration due to gravity, $g$.
Part (b): Terminal Speed \( v_T \)
After some time, the body reaches a constant speed, called the terminal velocity \( v_T \), where the drag force balances the gravitational force. At terminal velocity, the net force on the body is zero, meaning the downward gravitational force is exactly counteracted by the upward drag force.
The gravitational force is:
F_{\text{gravity}} = mg
The drag force at terminal velocity is:
D = bv_T^2
At terminal velocity, the drag force equals the gravitational force:
mg = bv_T^2
Solving for $v_T$, the terminal speed:
v_T = \sqrt{\frac{mg}{b}}
So, the terminal speed is:
v_T = \sqrt{\frac{mg}{b}}
Part (c): Downward Acceleration when \( v = \frac{v_T}{2} \)
Now we need to find the downward acceleration when the velocity of the body is \( v = \frac{v_T}{2} \). The net force on the body is the difference between the gravitational force and the drag force.
The gravitational force is still:
F_{\text{gravity}} = mg
The drag force when the velocity is $v = \frac{v_T}{2}$ is:
D = b \left( \frac{v_T}{2} \right)^2 = \frac{b v_T^2}{4}
Using the expression for terminal velocity $v_T = \sqrt{\frac{mg}{b}}$, we can substitute this into the drag force equation:
D = \frac{b \left( \frac{mg}{b} \right)}{4} = \frac{mg}{4}
The net force on the body at this velocity is:
F_{\text{net}} = mg - \frac{mg}{4} = \frac{3mg}{4}
Now, using Newton's second law, the acceleration is:
a = \frac{F_{\text{net}}}{m} = \frac{\frac{3mg}{4}}{m} = \frac{3g}{4}
So, the downward acceleration when $v = \frac{v_T}{2}$ is:
a = \frac{3g}{4}
Summary of Results
- Initial downward acceleration: \( a_{\text{initial}} = g \)
- Terminal speed: \( v_T = \sqrt{\frac{mg}{b}} \)
- Downward acceleration when \( v = \frac{v_T}{2} \): \( a = \frac{3g}{4} \)
