A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force q E, its velocity increases from 0 to $6 {\mathrm{ms}}^{-1}$ 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 s are respectively  

According to the question, 
For the time duration 0 < t < Is, the velocity increase from 0 to $6 {\mathrm{ms}}^{-1}$.
As the direction of field has been reversed, for 1 < t < 2 s, the velocity firstly decreases from $6 {\mathrm{ms}}^{-1}$ to 0.
Then, for 2 < t < 3 s; as the field strength is same; the magnitude of acceleration would be same, but velocity increases from 0 to $- 6 {\mathrm{ms}}^{-1}$.

Acceleration of the car,

               $\left|a\right|=\left|\frac{v-u}{t}\right|=\frac{6-0}{1}=6{ \mathrm{ms}}^{-2}$

The displacement of the particle is given as

              $s=ut+\frac{1}{2}a{t}^2$

For t = 0 to t = 1 s,

             $u=0,a=+6{ \mathrm{ms}}^{-2}\Rightarrow s_1=0+\frac{1}{2}\times 6\times (1{)}^2=3 \mathrm{m}$

For t = 1 s to t = 2 s,

             $\mathrm{u} = 6 {\mathrm{ms}}^{-1}, \mathrm{a} = -6 {\mathrm{ms}}^{-2}$

$\Rightarrow s_2=6\times 1-\frac{1}{2}\times 6\times (1{)}^2=6-3=3 \mathrm{m}$

For t = 2 s to t = 3 s,

             $u=0,a=-6{ \mathrm{ms}}^{-2}\Rightarrow s_3=0-\frac{1}{2}\times 6\times (1{)}^2=-3 \mathrm{m}$

$\therefore$  Net displacement,
            $\mathrm{s} = {\mathrm{s}}_1 + {\mathrm{s}}_2 + {\mathrm{s}}_3 = 3 \mathrm{m} + 3 \mathrm{m} - 3 \mathrm{m} = 3 \mathrm{m}$
Hence, average velocity

            $=\frac{\text{ net displacement }}{\text{ total time }}=\frac{3}{3}=1{ \mathrm{ms}}^{-1}$

Total distance travelled, d = 9 m
Hence, average speed

              $=\frac{\text{ total distance }}{\text{ total time }}=\frac{9}{3}=3{ \mathrm{ms}}^{-1}$

Alternate Method
Given condition can be represented through graph also as shown below.

$\therefore$   Displacement in three seconds
            = Area under the graph
            = Area of $∆ \mathrm{OAO}\text{'} + \mathrm{Area} \mathrm{of} ∆\mathrm{AO}\text{'}\mathrm{B} - \mathrm{Area} \mathrm{of} ∆\mathrm{BCD}$

                                      $=\frac{1}{2}\times 1\times 6+\frac{1}{2}\times 1\times 6-\frac{1}{2}\times 6\times 1=3 \mathrm{m}$

$\therefore$     Average velocity $=\frac{3}{3}=1{ \mathrm{ms}}^{-1}$

Total distance travelled, d = 9m

$\therefore$    Average velocity  $=\frac{9}{3}=3{ \mathrm{ms}}^{-1}$