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

Explanation

To solve the problem, we start with the differential equation given for the motion of the body:

\[ \frac{dv}{dt} = 6 - 3v \]

### Step 1: Rearranging and Integrating
We can rearrange the equation to separate the variables:

\[ \frac{dv}{6 - 3v} = dt \]

Next, we integrate both sides. The left side requires a substitution. We can rewrite it as:

\[ \int \frac{dv}{6 - 3v} = \int dt \]

### Step 2: Performing the Integration
The left side can be integrated using the natural logarithm:

\[ -\frac{1}{3} \ln|6 - 3v| = t + C \]

Where \( C \) is the constant of integration.

### Step 3: Applying Initial Conditions
At \( t = 0 \), the body is at rest, so \( v = 0 \):

\[ -\frac{1}{3} \ln|6 - 3(0)| = 0 + C \implies C = -\frac{1}{3} \ln(6) \]

Substituting \( C \) back into the equation gives:

\[ -\frac{1}{3} \ln|6 - 3v| = t - \frac{1}{3} \ln(6) \]

### Step 4: Solving for \( v \)
Rearranging gives:

\[ \ln|6 - 3v| = -3t + \ln(6) \]

Exponentiating both sides:

\[ 6 - 3v = 6e^{-3t} \]

Solving for \( v \):

\[ 3v = 6 - 6e^{-3t} \implies v = 2(1 - e^{-3t}) \]

### Step 5: Finding Terminal Speed
The terminal speed occurs as \( t \to \infty \):

\[ v = 2(1 - e^{-3t}) \to 2(1 - 0) = 2 \text{ m/s} \]

### Step 6: Finding Initial Acceleration
To find the initial acceleration, we differentiate \( v \):

\[ a = \frac{dv}{dt} = 6 - 3v \]

At \( t = 0 \), \( v = 0 \):

\[ a = 6 - 3(0) = 6 \text{ m/s}^2 \]

### Step 7: Finding Speed When Acceleration is Half
The initial acceleration is \( 6 \text{ m/s}^2 \), so half of that is \( 3 \text{ m/s}^2 \):

\[ 3 = 6 - 3v \implies 3v = 3 \implies v = 1 \text{ m/s} \]

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