A toy car with charge q is on a frictionless horizontal plane surface under the influence of a uniform electric field \( \vec{E} \). Due to the force \( q\vec{E} \), its velocity increases from 0 to \( 6\text{ m/s} \) in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

Acceleration \( a = \frac{6}{1} = 6\text{ m/s}^2 \). Distance in first second \( x_1 = \frac{1}{2}(6)(1)^2 = 3\text{ m} \). At \( t=1\text{s} \), velocity \( v=6\text{ m/s} \). For next 2 seconds, displacement \( x_2 = 6(2) - \frac{1}{2}(6)(2)^2 = 12 - 12 = 0 \). Total displacement = \( 3 + 0 = 3\text{ m} \). Average velocity = \( \frac{3}{3} = 1\text{ m/s} \). Total distance = \( 3 + |6(2) - \frac{1}{2}(6)(4)| \) wait, distance turned = \( 3 + 6 + 6 = 15\text{ m} \). Total distance = \( 3 + 6 + 6 = 15\text{ m} \implies \) Average speed = \( \frac{15}{3} = 5\text{ m/s} \) (Re-evaluating options: Option 2 gives \( 1\text{ m/s, } 3\text{ m/s} \)... wait, calculation check: distance traveled in deceleration: comes to rest at \( t = 1 + 1 = 2\text{s} \), moves back 3m by \( t=3\text{s} \). Total distance = \( 3 + 3 + 3 = 9\text{ m} \). Average speed = \( 9/3 = 3\text{ m/s} \). Average velocity = \( 3/3 = 1\text{ m/s} \)).