Displacement is the integral of velocity with respect to time. $\Delta s = \int_{t_1}^{t_2} v(t) \, dt$

Substituting the given equation and limits: $\Delta s = \int_{1}^{3} (3t^2 - 2t + 1) \, dt$

Integrating term by term using the power rule $\int t^n \, dt = \frac{t^{n+1}}{n+1}$: $\Delta s = \left[ t^3 - t^2 + t \right]_{1}^{3}$

Now, substitute the upper and lower limits: At $t = 3$: $3^3 - 3^2 + 3 = 27 - 9 + 3 = 21$

At $t = 1$: $1^3 - 1^2 + 1 = 1$

Subtracting the lower limit value from the upper limit value: $\Delta s = 21 - 1 = 20\text{ m}$

• Option A is correct because the definite integration yields exactly $20\text{ m}$.
• Option B, C, and D are incorrect due to common mathematical errors, such as forgetting to subtract the lower limit ($21$ m) or incorrect differentiation instead of integration ($v'(t) = 6t - 2$).