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 step by step, we will analyze the given equation and find the speed of the body as a function of time.

### Step 1: Understand the given equation
The motion of the body is described by the equation:
\[ \frac{dv}{dt} = 6 - 3v \]
where \( v \) is the speed in \( m/s \) and \( t \) is time in seconds.

### Step 2: Rearrange the equation
We can rearrange the equation to isolate \( dv \):
\[ \frac{dv}{6 - 3v} = dt \]

### Step 3: Integrate both sides
Now we will integrate both sides. The left side will require a logarithmic integration:
\[ \int \frac{1}{6 - 3v} dv = \int dt \]
The integral of the left side can be solved using the substitution method:
\[ \int \frac{1}{6 - 3v} dv = -\frac{1}{3} \ln |6 - 3v| + C_1 \]
The integral of the right side is simply:
\[ t + C_2 \]

### Step 4: Combine the results
Setting the two integrals equal gives us:
\[ -\frac{1}{3} \ln |6 - 3v| = t + C \]
where \( C = C_2 - C_1 \).

### Step 5: Solve for \( v \)
To solve for \( v \), we can exponentiate both sides:
\[ |6 - 3v| = e^{-3(t + C)} = e^{-3t} e^{-3C} \]
Let \( K = e^{-3C} \), then:
\[ 6 - 3v = K e^{-3t} \]
Now, solving for \( v \):
\[ 3v = 6 - K e^{-3t} \]
\[ v = 2 - \frac{K}{3} e^{-3t} \]

### Step 6: Apply the initial condition
The problem states that the body is at rest at \( t = 0 \), which means \( v(0) = 0 \):
\[ 0 = 2 - \frac{K}{3} e^{0} \]
\[ 0 = 2 - \frac{K}{3} \]
This implies:
\[ \frac{K}{3} = 2 \implies K = 6 \]

### Step 7: Substitute \( K \) back into the equation for \( v \)
Substituting \( K \) back into the equation for \( v \):
\[ v = 2 - \frac{6}{3} e^{-3t} \]
\[ v = 2 - 2 e^{-3t} \]

### Final Result
Thus, the speed of the body as a function of time is:
\[ v(t) = 2 - 2 e^{-3t} \]