A car starting from rest accelerates uniformly at $2$ m/s². The distance covered in $5$ seconds is:

Using $s = u t + \dfrac{1}{2} a t^2$ with $u = 0$, $a = 2$ m/s², $t = 5$ s: $s = 0 + \dfrac{1}{2} \times 2 \times 25 = 25$ m. Alternative: final velocity $v = at = 10$ m/s. Average velocity (from rest at constant acceleration) is $\bar v = \dfrac{0 + 10}{2} = 5$ m/s. Distance $= \bar v \cdot t = 5 \times 5 = 25$ m ✓. Option B (20 m) is the trap if you compute $v \times t = 10 \times 5 / 2.5$ or otherwise mishandle the factor of $\dfrac{1}{2}$.