7. Given that $$ \frac{d v}{d t}=3-2 t $$ and $$ v=2 $$ when $$ t=1 $$, find the value of $$ v $$ when $$ t=2 $$.

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To find the value of $$ v $$ when $$ t = 2 $$, we'll start by integrating the given differential equation and then apply the initial condition.

**Given:**
$$
\frac{dv}{dt} = 3 - 2t
$$
$$
v(1) = 2
$$

**Step 1: Integrate the Differential Equation**

Integrate $$ \frac{dv}{dt} $$ with respect to $$ t $$:
$$
v(t) = \int (3 - 2t) \, dt
$$
$$
v(t) = 3t - t^2 + C
$$
where $$ C $$ is the constant of integration.

**Step 2: Apply the Initial Condition**

Use the initial condition $$ v(1) = 2 $$ to find $$ C $$:
$$
2 = 3(1) - (1)^2 + C
$$
$$
2 = 3 - 1 + C
$$
$$
2 = 2 + C
$$
$$
C = 0
$$

**Step 3: Write the Particular Solution**

With $$ C = 0 $$, the solution simplifies to:
$$
v(t) = 3t - t^2
$$

**Step 4: Find $$ v(2) $$**

Substitute $$ t = 2 $$ into the equation:
$$
v(2) = 3(2) - (2)^2
$$
$$
v(2) = 6 - 4
$$
$$
v(2) = 2
$$

**Answer:**  
When $$ t = 2 $$, the value of $$ v $$ is **2**.
When $$ t = 2 $$, $$ v = 2 $$.

7. Given that \( \frac{d v}{d t}=3-2 t \) and \( v=2 \) when \( t=1 \), find the value of \( v \) when \( t=2 \).

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