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 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 ms-1.As the direction of field has been reversed, for 1 < t < 2 s, the velocity firstly decreases from 6 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 ms-1.Acceleration of the car,               |a|=v-ut=6-01=6 ms-2The displacement of the particle is given as              s=ut+12at2For t = 0 to t = 1 s,             u=0,a=+6 ms-2⇒s1=0+12×6×(1)2=3 mFor t = 1 s to t = 2 s,             u = 6 ms-1,  a = -6 ms-2⇒      s2=6×1-12×6×(1)2=6-3=3 mFor t = 2 s to t = 3 s,             u=0,a=-6 ms-2⇒s3=0-12×6×(1)2=-3 m∴  Net displacement,            s = s1 + s2 + s3 = 3 m + 3 m - 3 m = 3 mHence, average velocity            = net displacement  total time =33=1 ms-1Total distance travelled, d = 9 mHence, average speed              = total distance  total time =93=3 ms-1Alternate MethodGiven condition can be represented through graph also as shown below.∴   Displacement in three seconds            = Area under the graph            = Area of ∆ OAO' + Area of ∆AO'B - Area of ∆BCD                                      =12×1×6+12×1×6-12×6×1=3 m∴     Average velocity =33=1 ms-1Total distance travelled, d = 9m∴    Average velocity  =93=3 ms-1