A toy car with charge `q` moves 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 m//s` in one second duration. At that instant the direction of 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.

Options

A. 1 m/s, 3.5 m/s
B. 1 m/s, 3 m/s
C. 2 m/s , 4 m/s
D. 1.5 m/s , 3 m/s

Answer

B

Explanation

When field is reversed then car comes to rest in one second and in next one secund it returns to the point where it was at t =1s.

Car acquires a speed of 6 m/s in one second hence
v=u+at `rArr` 6=0 + a x 1 `rArr a=6 m//s^2`
For t=0 to t=1 s , (initial velocity is zero)
`S_1=0+1/2xx6(1)^2`=3m …(i)
For t=1 s to t=2 s, (initial velocity is 6 m/s)
`S_2=6-1/2xx6(1)^2`=3m ...(ii)
For t=2 s to t=3 s , (particle starts from rest in reversed direction)
`S_3=0-1/2xx6(1)^2`=-3 m ....(iii)
Total displacement `S=S_1+S_2+S_3=3` m
Average velocity =`"Total displacement "/"Total time"=3/3 =1 ms^(-1)`
Total distance travelled =9 m
Average speed = `"Total distance "/"Total time" =9/3 =3 ms^(-1)`