To find the distance travelled by the particle between 1 s and 2 s, we need to integrate the velocity function.$\\$Given velocity function:$\\ v = At + Bt^2 \\$The distance $s$ travelled between time $t_1$ and $t_2$ is given by the integral of velocity:$\\ s = \int_{t_1}^{t_2} v \, dt \\$Substitute the given velocity function:$\\ s = \int_{1}^{2} (At + Bt^2) \, dt \\$Split the integral:$\\ s = \int_{1}^{2} At \, dt + \int_{1}^{2} Bt^2 \, dt \\$Integrate each term separately:$\\ \int At \, dt = \frac{A}{2} t^2 \Big|_{1}^{2} \\ \int Bt^2 \, dt = \frac{B}{3} t^3 \Big|_{1}^{2} \\$Evaluate the integrals:$\\ \frac{A}{2} t^2 \Big|_{1}^{2} = \frac{A}{2} (2^2) - \frac{A}{2} (1^2) \\ = \frac{A}{2} (4) - \frac{A}{2} (1) \\ = 2A - \frac{A}{2} \\ = \frac{4A}{2} - \frac{A}{2} \\ = \frac{3A}{2} \\ \frac{B}{3} t^3 \Big|_{1}^{2} = \frac{B}{3} (2^3) - \frac{B}{3} (1^3) \\ = \frac{B}{3} (8) - \frac{B}{3} (1) \\ = \frac{8B}{3} - \frac{B}{3} \\ = \frac{7B}{3} \\$Add the results of the integrals to find the total distance:$\\ s = \frac{3A}{2} + \frac{7B}{3} \\$Therefore, the distance travelled by the particle between 1 s and 2 s is $\frac{3A}{2} + \frac{7B}{3}. \\$This corresponds to Option 1.