The Question is from neet-physics electrostatic-charges-and-fields , and it was created by the author VENKATESWARARAO

Question

A toy car with charge ${\rm{q}}$ moves on a frictionless horizontal plane surface under the influence of a uniform electric field $\overrightarrow {\rm{E}} $ . Due to the force ${\rm{q}}$, $\overrightarrow {\rm{E}} $ its velocity increases from ${\rm{0}}\,\,{\rm{to}}\,\,{\rm{6 m}}\,{\rm{/}}\,{\rm{s}}$ in one second duration. At that instant the direction of the filed is reversed. The care 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 ${\rm{0}}\,\,{\rm{to}}\,\,{\rm{3}}$ seconds are respectively

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
	**Primary Concept**: Kinematics and Uniform Acceleration
	**Related Topics**: Average Velocity, Average Speed, Constant Acceleration
	**Prerequisites**: Equations of motion, vector addition
## SOLUTION PATHWAY
**Given Information**:
	* Initial velocity, $u = 0$ m/s
	* Velocity after 1 second, $v_1 = 6$ m/s
	* Field reversed after 1 second.
	* Motion continues for 2 more seconds.
**Method & Steps**:
	1. **Calculate acceleration in the first second**:
		Reasoning: Since the field is uniform, the acceleration is constant.
		Formula/Rule: $v = u + at$
		Working: $6 = 0 + a(1) \Rightarrow a = 6$ m/s²
	2. **Calculate displacement in the first second**:
		Reasoning: Using equations of motion
		Formula/Rule: $s = ut + \frac{1}{2}at^2$
		Working: $s_1 = (0)(1) + \frac{1}{2}(6)(1)^2 = 3$ m
	3. **Calculate the velocity and displacement after 2 more seconds (field reversed):**
		Reasoning: Acceleration is now -6 m/s²
		Formula/Rule: $v = u + at$ and $s = ut + \frac{1}{2}at^2$
		Working:  $v_3 = 6 + (-6)(2) = -6$ m/s and  $s_2 = (6)(2) + \frac{1}{2}(-6)(2)^2 = 0$ m
	4. **Calculate total displacement:**
		Working: $s_{total} = s_1 + s_2 = 3 + 0 = 3$ m
	5. **Calculate average velocity:**
		Formula/Rule: Average velocity $= \frac{	ext{Total Displacement}}{	ext{Total Time}}$
		Working: Average velocity $= \frac{3}{3} = 1$ m/s
	6. **Calculate total distance:**
		Reasoning: Distance is a scalar, so we consider magnitudes only.
		Working: In the first second, distance = 3 m. In the next two seconds, the car goes to 0 velocity and comes back to the initial position in the reversed field, covering 6 m in one direction and 6 m back (since $v = 6$ m/s at $t=1$s and $v=0$ at $t=2$s and then $v=-6$m/s at $t=3$s). Therefore distance covered in 2 seconds is $2* ( \frac{1}{2} * 6 * 1) = 6$ m. Total distance is 3+6 = 9m.
	7. **Calculate average speed:**
		Formula/Rule: Average speed $= \frac{	ext{Total Distance}}{	ext{Total Time}}$
		Working: Average speed $= \frac{9}{3} = 3$ m/s

## OPTION ANALYSIS
[A] :x: Incorrect. 
[B] :x: Incorrect
[C] :x: Incorrect
[D] ✓: Correct. Average velocity = 1 m/s and Average speed = 3 m/s
## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing distance and displacement: Remember displacement is a vector while distance is a scalar.
		2. Incorrectly applying signs to acceleration: When the field reverses, the acceleration changes sign.
	**Quick Tips**:
		1. Visualize the motion: Drawing a velocity-time graph can help understand the motion.
		2. Remember the definitions: Average velocity is displacement over time, and average speed is distance over time.

VENKATESWARARAO · Answer

${\rm{1}}{\rm{.5}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s,}}\,\,{\rm{3}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s}}$

VENKATESWARARAO · Answer

${\rm{2}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s,}}\,\,{\rm{4}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s\;\;\;\;\;\;}}$

VENKATESWARARAO · Answer

${\rm{1}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s,}}\,\,{\rm{3.5}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s}}$

VENKATESWARARAO · Answer

The correct Option is "${\rm{1}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s,}}\,\,{\rm{3}}\,\,{\rm{m}}\,{\rm{/}}\,{\rm{s}}$"

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