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

Answer

To solve the problem step by step, we will analyze the motion of the toy car in three distinct time intervals: from 0 to 1 second, from 1 to 2 seconds, and from 2 to 3 seconds.

### Step 1: Analyze the motion from 0 to 1 second
- **Initial velocity (u)**: 0 m/s (the car starts from rest)
- **Final velocity (v)**: 6 m/s (after 1 second)
- **Time (t)**: 1 second

Using the formula for acceleration:
\[ a = \frac{v - u}{t} = \frac{6 - 0}{1} = 6 \, \text{m/s}^2 \]

Now, we can calculate the distance traveled (s1) during this interval using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s_1 = 0 \cdot 1 + \frac{1}{2} \cdot 6 \cdot (1)^2 = 3 \, \text{m} \]

### Step 2: Analyze the motion from 1 to 2 seconds
- **Initial velocity (u)**: 6 m/s (the velocity at the end of the first second)
- **Final velocity (v)**: To find this, we need to calculate the effect of the reversed electric field.
- **Time (t)**: 1 second

Since the electric field is reversed, the acceleration will be negative. The acceleration (a) is equal to -6 m/s² (the same magnitude but opposite direction).

Using the formula for final velocity:
\[ v = u + at = 6 + (-6) \cdot 1 = 0 \, \text{m/s} \]

Now, we can calculate the distance traveled (s2) during this interval:
\[ s_2 = ut + \frac{1}{2} a t^2 = 6 \cdot 1 + \frac{1}{2} \cdot (-6) \cdot (1)^2 = 6 - 3 = 3 \, \text{m} \]

### Step 3: Analyze the motion from 2 to 3 seconds
- **Initial velocity (u)**: 0 m/s (the velocity at the end of the second second)
- **Final velocity (v)**: To find this, we again consider the reversed field.
- **Time (t)**: 1 second

The acceleration remains -6 m/s².

Using the formula for final velocity:
\[ v = u + at = 0 + (-6) \cdot 1 = -6 \, \text{m/s} \]

Now, we can calculate the distance traveled (s3) during this interval:
\[ s_3 = ut + \frac{1}{2} a t^2 = 0 \cdot 1 + \frac{1}{2} \cdot (-6) \cdot (1)^2 = -3 \, \text{m} \]

### Step 4: Calculate total distance and displacement
- **Total distance traveled**:
\[ \text{Total distance} = s_1 + |s_2| + |s_3| = 3 + 3 + 3 = 9 \, \text{m} \]

- **Total displacement**:
\[ \text{Displacement} = s_1 + s_2 + s_3 = 3 + 3 - 3 = 3 \, \text{m} \]

### Step 5: Calculate average velocity and average speed
- **Average speed**:
\[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{9 \, \text{m}}{3 \, \text{s}} = 3 \, \text{m/s} \]

- **Average velocity**:
\[ \text{Average velocity} = \frac{\text{Displacement}}{\text{Total time}} = \frac{3 \, \text{m}}{3 \, \text{s}} = 1 \, \text{m/s} \]

### Final Answer:
- Average velocity: **1 m/s**
- Average speed: **3 m/s**