The motion of a body is given by the equation `dv//dt=6-3v`, where v is in m//s. If the body was at rest at `t=0`
(i) the terminal speed is `2 m//s`
(ii) the magnitude of the initial acceleration is `6 m//s^(2)`
(iii) The speed varies with time as `v=2(1-e^(-3t)) m//s`
(iv) The speed is `1 m//s`, when the acceleration is half initial value

Options

A. (i), (ii)
B. (ii), (iii),(iv)
C. (i), (ii), (iii)
D. All

Explanation

To solve the problem step by step, we will analyze the given differential equation and verify each of the statements provided in the question.

### Step 1: Understand the given differential equation
The motion of the body is described by the equation:
\[ \frac{dv}{dt} = 6 - 3v \]
where \( v \) is the velocity of the body in meters per second.

### Step 2: Find the terminal speed
The terminal speed occurs when acceleration becomes zero, i.e., when \(\frac{dv}{dt} = 0\).
Setting the equation to zero:
\[ 0 = 6 - 3v \]
Solving for \( v \):
\[ 3v = 6 \implies v = 2 \, \text{m/s} \]
Thus, the terminal speed is \( 2 \, \text{m/s} \).

### Step 3: Calculate the initial acceleration
The initial acceleration can be found by substituting \( v = 0 \) into the acceleration equation:
\[ \frac{dv}{dt} = 6 - 3(0) = 6 \, \text{m/s}^2 \]
Thus, the magnitude of the initial acceleration is \( 6 \, \text{m/s}^2 \).

### Step 4: Solve the differential equation
We can rearrange the equation for integration:
\[ \frac{dv}{6 - 3v} = dt \]
Integrating both sides:
\[ \int \frac{1}{6 - 3v} \, dv = \int dt \]
The left side can be integrated using the substitution \( u = 6 - 3v \), leading to:
\[ -\frac{1}{3} \ln |6 - 3v| = t + C \]
Exponentiating both sides gives:
\[ 6 - 3v = Ce^{-3t} \]
Solving for \( v \):
\[ 3v = 6 - Ce^{-3t} \implies v = 2 - \frac{C}{3} e^{-3t} \]
At \( t = 0 \), the body is at rest (\( v = 0 \)):
\[ 0 = 2 - \frac{C}{3} \implies C = 6 \]
Thus, the velocity function is:
\[ v = 2 - 2e^{-3t} \]
Rearranging gives:
\[ v = 2(1 - e^{-3t}) \, \text{m/s} \]

### Step 5: Check when acceleration is half of the initial value
The initial acceleration is \( 6 \, \text{m/s}^2 \), so half of that is \( 3 \, \text{m/s}^2 \).
Setting the acceleration equation equal to \( 3 \):
\[ 6 - 3v = 3 \]
Solving for \( v \):
\[ 3v = 3 \implies v = 1 \, \text{m/s} \]

### Summary of Results
1. The terminal speed is \( 2 \, \text{m/s} \) (True).
2. The magnitude of the initial acceleration is \( 6 \, \text{m/s}^2 \) (True).
3. The speed varies with time as \( v = 2(1 - e^{-3t}) \, \text{m/s} \) (True).
4. The speed is \( 1 \, \text{m/s} \) when the acceleration is half of the initial value (True).

### Final Conclusion
All statements provided in the question are correct.