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

Explanation

To solve the problem, we will break it down into steps to find the average velocity and average speed of the toy car over the time interval from 0 to 3 seconds.

### Step 1: Analyze the motion from 0 to 1 second
- The car starts from rest (initial velocity \( u = 0 \, \text{m/s} \)) and accelerates to \( 6 \, \text{m/s} \) in 1 second.
- We can use the formula for displacement under uniform acceleration:
\[ s_1 = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \), \( t = 1 \, \text{s} \), and we need to find \( a \).

- The final velocity \( v = 6 \, \text{m/s} \) can be related to acceleration:
\[ v = u + at \implies 6 = 0 + a \cdot 1 \implies a = 6 \, \text{m/s}^2 \]

- Now substituting \( a \) back into the displacement formula:
\[ s_1 = 0 \cdot 1 + \frac{1}{2} \cdot 6 \cdot (1)^2 = 3 \, \text{m} \]

### Step 2: Analyze the motion from 1 to 3 seconds
- At \( t = 1 \, \text{s} \), the velocity is \( 6 \, \text{m/s} \). The electric field is reversed, so the car will decelerate.
- The acceleration will be \( -6 \, \text{m/s}^2 \) (the same magnitude but opposite direction).

- From \( t = 1 \) to \( t = 3 \) seconds (2 seconds), we can find the displacement:
- The initial velocity for this segment is \( 6 \, \text{m/s} \) and the time is \( t = 2 \, \text{s} \).

- Using the formula again:
\[ s_2 = ut + \frac{1}{2} a t^2 \]
Here, \( u = 6 \, \text{m/s} \), \( a = -6 \, \text{m/s}^2 \), and \( t = 2 \, \text{s} \):
\[ s_2 = 6 \cdot 2 + \frac{1}{2} \cdot (-6) \cdot (2)^2 = 12 - 12 = 0 \, \text{m} \]

### Step 3: Calculate total displacement and average velocity
- The total displacement over the entire 3 seconds:
\[ \text{Total Displacement} = s_1 + s_2 = 3 + 0 = 3 \, \text{m} \]

- The average velocity \( v_{avg} \) is given by:
\[ v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{3 \, \text{m}}{3 \, \text{s}} = 1 \, \text{m/s} \]

### Step 4: Calculate total distance and average speed
- The total distance covered is the distance from \( 0 \) to \( 1 \) second plus the distance from \( 1 \) to \( 3 \) seconds:
- Distance from \( 0 \) to \( 1 \) second: \( 3 \, \text{m} \)
- Distance from \( 1 \) to \( 3 \) seconds: The car moves at \( 6 \, \text{m/s} \) for \( 1 \, \text{s} \) and then decelerates to \( 0 \, \text{m/s} \) over the next \( 1 \, \text{s} \), covering \( 6 \, \text{m} \) in the first second and \( 0 \, \text{m} \) in the second second.

- Therefore, the total distance is:
\[ \text{Total Distance} = 3 + 6 = 9 \, \text{m} \]

- The average speed \( v_{avg, speed} \) is given by:
\[ v_{avg, speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{9 \, \text{m}}{3 \, \text{s}} = 3 \, \text{m/s} \]

### Final Answers
- Average Velocity: \( 1 \, \text{m/s} \)
- Average Speed: \( 3 \, \text{m/s} \)