The motion of a body is given by the equation d`nu`/dt = 6 - 3`nu` where `nu` is the speed in `m s^(-1)` and t is time in s. The body is at rest at t = 0. The speed varies with time as

Options

A. `nu = (1 - e^(-3t))`
B. `nu = 2(1 - e^(-3t))`
C. `nu = 1 + e^(-2t)`
D. `nu = 2(1 + e^(-2t))`

Explanation

To solve the problem where the motion of a body is given by the equation \(\frac{dv}{dt} = 6 - 3v\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ \frac{dv}{dt} = 6 - 3v \]
We can rearrange this equation to separate variables:
\[ \frac{dv}{6 - 3v} = dt \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side involves a logarithmic integration:
\[ \int \frac{dv}{6 - 3v} = \int dt \]
The integral on the left can be solved using the substitution method or recognizing it as a standard integral.

### Step 3: Solving the Left Integral
The integral of \(\frac{1}{6 - 3v}\) can be computed as follows:
\[ \int \frac{1}{6 - 3v} dv = -\frac{1}{3} \ln |6 - 3v| + C \]
where \(C\) is the constant of integration. The integral on the right side is simply:
\[ \int dt = t + C' \]

### Step 4: Equating the Integrals
Setting the two integrals equal gives:
\[ -\frac{1}{3} \ln |6 - 3v| = t + C' \]

### Step 5: Applying Initial Conditions
Since the body is at rest at \(t = 0\), we have \(v(0) = 0\). We substitute \(t = 0\) and \(v = 0\) into the equation:
\[ -\frac{1}{3} \ln |6 - 3(0)| = 0 + C' \]
This simplifies to:
\[ -\frac{1}{3} \ln 6 = C' \]

### Step 6: Substituting Back to Find \(v\)
Now we substitute \(C'\) back into our equation:
\[ -\frac{1}{3} \ln |6 - 3v| = t - \frac{1}{3} \ln 6 \]
Multiplying through by -3 gives:
\[ \ln |6 - 3v| = -3t + \ln 6 \]

### Step 7: Exponentiating Both Sides
Exponentiating both sides to eliminate the logarithm:
\[ |6 - 3v| = e^{-3t} \cdot 6 \]

### Step 8: Solving for \(v\)
We can drop the absolute value since \(6 - 3v\) will be positive for our range of \(v\):
\[ 6 - 3v = 6e^{-3t} \]
Rearranging gives:
\[ 3v = 6 - 6e^{-3t} \]
\[ v = 2(1 - e^{-3t}) \]

### Final Answer
Thus, the speed \(v\) varies with time as:
\[ v(t) = 2(1 - e^{-3t}) \]