1. The distance travelled between $t=1$ s and $t=2$ s is obtained by integrating the velocity over that interval:

$s = \int_{1}^{2} v\,dt = \int_{1}^{2} (\alpha t + \beta t^2)\,dt$

2. Integrate term by term:

$\int \alpha t\,dt = \alpha \cdot \frac{t^2}{2}$

$\int \beta t^2\,dt = \beta \cdot \frac{t^3}{3}$

So,

$s = \left[\frac{\alpha t^2}{2} + \frac{\beta t^3}{3}\right]_{1}^{2}$

3. Substitute the limits:

At $t=2$:

$\frac{\alpha(2)^2}{2} + \frac{\beta(2)^3}{3} = \frac{4\alpha}{2} + \frac{8\beta}{3} = 2\alpha + \frac{8\beta}{3}$

At $t=1$:

$\frac{\alpha(1)^2}{2} + \frac{\beta(1)^3}{3} = \frac{\alpha}{2} + \frac{\beta}{3}$

4. Subtract:

$s = \left(2\alpha + \frac{8\beta}{3}\right) - \left(\frac{\alpha}{2} + \frac{\beta}{3}\right)$

$s = \frac{3\alpha}{2} + \frac{7\beta}{3}$

5. Match with the options:

$\boxed{\frac{3}{2}\alpha + \frac{7}{3}\beta}$

This corresponds to Option B.