The given equation is dv/dt = 6 - 3v, where v represents the speed of the body in m/s and t represents time in seconds. We need to find the equation for v in terms of t.

To solve this differential equation, we can use the method of separation of variables.

1. Separation of Variables:
- Move the term involving v to one side and the term involving t to the other side:
dv/(6 - 3v) = dt

2. Integration:
- Integrate both sides with respect to their respective variables:
∫(1/(6 - 3v)) dv = ∫dt

- On the left side, we can apply a substitution to simplify the integral. Let u = 6 - 3v, then du = -3dv:
-1/3 ∫(1/u) du = ∫dt

- Integrating both sides gives:
-1/3 ln|u| = t + C1

- Simplifying the natural logarithm:
ln|u| = -3t + C1

3. Solve for u:
- Taking the exponential of both sides eliminates the natural logarithm:
|u| = e^(-3t + C1)

- Considering the absolute value, we can write:
u = ±e^(-3t + C1)

4. Solve for v:
- Recall that u = 6 - 3v. Substituting this into the equation gives:
6 - 3v = ±e^(-3t + C1)

- Rearranging the equation yields:
3v = 6 - e^(-3t + C1)

- Dividing both sides by 3:
v = 2 - (1/3)e^(-3t + C1)

5. Determine the constant:
- We know that the body is at rest at t = 0, which means v = 0. Substituting these values into the equation allows us to determine the constant:
0 = 2 - (1/3)e^(C1)

- Solving for C1:
(1/3)e^(C1) = 2

- Taking the natural logarithm of both sides:
ln((1/3)e^(C1)) = ln(2)

- Simplifying:
C1 + ln(1/3) = ln(2)
C1 = ln(2) - ln(1/3)
C1 = ln(2/1/3)
C1 = ln(6)

6. Final solution:
- Substituting the value of C1 back into the equation for v:
v = 2 - (1/3)e^(-3t + ln(6))

- Simplifying:
v = 2 - (1/3)e^(ln(6) - 3t)
v = 2 - (1/3)(6e^(-3t))
v = 2 - 2e^(-3t)
v = 2[1 - e^(-3t)]

Therefore, the speed of the body as a function of time is v = 2[1 - e^(-3t)].