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

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**.