To find the acceleration of the particle at $t = 2$ s, we need to determine the second derivatives of the position functions $x(t)$ and $y(t). \\$Given:$\\ x = 5t - 2t^2 \\ y = 10t \\$First, we find the velocity components by differentiating the position functions with respect to time $t. \\$Velocity in the $x$-direction:$\\ v_x = \frac{dx}{dt} = \frac{d}{dt}(5t - 2t^2) = 5 - 4t \\$Velocity in the $y$-direction:$\\ v_y = \frac{dy}{dt} = \frac{d}{dt}(10t) = 10 \\$Next, we find the acceleration components by differentiating the velocity functions with respect to time $t. \\$Acceleration in the $x$-direction:$\\ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(5 - 4t) = -4 \\$Acceleration in the $y$-direction:$\\ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(10) = 0 \\$The total acceleration of the particle is given by the vector sum of the acceleration components:$\\ a = \sqrt{a_x^2 + a_y^2} \\$Substitute the values:$\\ a = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ m/s$^2 \\$Since the acceleration in the $x$-direction is negative, the correct answer is $-4$ m/s$^2. \\$Therefore, the acceleration of the particle at $t = 2$ s is $-4$ m/s$^2,$ which corresponds to Option 2.