Here's the solution for the instantaneous velocity: The equation for the acceleration of a particle is given by $a = 2 + 2t$. We need to find the instantaneous velocity after $4$ seconds. Step 1: Integrate the acceleration function to find the velocity function. Velocity $v(t)$ is the integral of acceleration $a(t)$ with respect to time $t$: $$v(t) = \int a(t)\,dt$$ Given $a(t) = 2 + 2t$: $$v(t) = \int (2 + 2t)\,dt$$ $$v(t) = \int 2\,dt + \int 2t\,dt$$ $$v(t) = 2t + 2\left(\frac{t^2}{2}\right) + C$$ $$v(t) = 2t + t^2 + C$$ Here, $C$ is the constant of integration, representing the initial velocity $v(0)$. Since no initial velocity is given, we assume the particle starts from rest, meaning $v(0) = 0$. Substituting $t=0$ and $v(0)=0$: $$0 = 2(0) + (0)^2 + C$$ $$C = 0$$ So, the velocity function is: $$v(t) = 2t + t^2$$ Step 2: Calculate the instantaneous velocity at $t = 4\text{s}$. Substitute $t = 4$ into the velocity function: $$v(4) = 2(4) + (4)^2$$ $$v(4) = 8 + 16$$ $$v(4) = 24\text{m/s}$$ Comparing this result with the given options: A. $8\text{m/s}$ B. $32\text{m/s}$ C. $24\text{m/s}$ D. $42\text{m/s}$ The calculated instantaneous velocity matches option C. The final answer is $\boxed{\text{C. 24 m/s}}$.