To solve this problem, we need to analyze the motion of the toy car in two phases:$\\$Phase 1: From $t = 0$ to $t = 1$ second (initial acceleration)$\\$Phase 2: From $t = 1$ to $t = 3$ seconds (after reversing the field)$\\$Given:$\\$• Initial velocity $u = 0$ m/s$\\$• Final velocity after 1 second $v = 6$ m/s$\\$The acceleration $a$ can be calculated using the formula:$\\ v = u + at \\ 6 = 0 + a \cdot 1 \\ a = 6$ m/s$^2 \\$Phase 1:$\\$The distance covered in the first second $(s_1)$ is given by:$\\ s_1 = ut + \frac{1}{2}at^2 \\ s_1 = 0 \cdot 1 + \frac{1}{2} \cdot 6 \cdot 1^2 \\ s_1 = 3$ m$\\$Phase 2:$\\$At $t = 1$ second, the electric field is reversed, so the acceleration becomes $-6$ m/s$^2. \\$The velocity at $t = 2$ seconds is:$\\ v_2 = v + (-a) \cdot 1 \\ v_2 = 6 - 6 \cdot 1 \\ v_2 = 0$ m/s$\\$The velocity at $t = 3$ seconds is:$\\ v_3 = v_2 + (-a) \cdot 1 \\ v_3 = 0 - 6 \cdot 1 \\ v_3 = -6$ m/s$\\$The distance covered from $t = 1$ to $t = 2$ seconds $(s_2)$ is:$\\ s_2 = v \cdot t + \frac{1}{2}(-a)t^2 \\ s_2 = 6 \cdot 1 + \frac{1}{2}(-6) \cdot 1^2 \\ s_2 = 6 - 3 \\ s_2 = 3$ m$\\$The distance covered from $t = 2$ to $t = 3$ seconds $(s_3)$ is:$\\ s_3 = v_2 \cdot t + \frac{1}{2}(-a)t^2 \\ s_3 = 0 \cdot 1 + \frac{1}{2}(-6) \cdot 1^2 \\ s_3 = -3$ m$\\$Total displacement $(s_{total})$ from $t = 0$ to $t = 3$ seconds is:$\\ s_{total} = s_1 + s_2 + s_3 \\ s_{total} = 3 + 3 - 3 \\ s_{total} = 3$ m$\\$Average velocity $(v_{avg})$ is given by:$\\ v_{avg} = \frac{s_{total}}{t_{total}} \\ v_{avg} = \frac{3}{3} \\ v_{avg} = 1$ m/s$\\$Total distance traveled $(d_{total})$ is:$\\ d_{total} = |s_1| + |s_2| + |s_3| \\ d_{total} = 3 + 3 + 3 \\ d_{total} = 9$ m$\\$Average speed $(v_{speed})$ is given by:$\\ v_{speed} = \frac{d_{total}}{t_{total}} \\ v_{speed} = \frac{9}{3} \\ v_{speed} = 3$ m/s$\\$Therefore, the average velocity and average speed are $1$ m/s and $3$ m/s respectively.$\\$This corresponds to Option 2.