The Question is from neet-physics electrostatic-charges-and-fields , and it was created by the author Mr. ANIL KUMAR.G

Question

A toy car with charge $q$ moves on a frictionless horizontal plane under a uniform electric field $\overrightarrow{E}$. Its velocity increases from 0 to $6 \, m/s$ in one second due to the force $q\overrightarrow{E}$.  If the field's direction is reversed at that instant, and the car continues for two more seconds, what are its average velocity and average speed over the three seconds?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Kinematics, Uniformly Accelerated Motion, Average Velocity, Average Speed
	**Related Topics**: Electric Force, Newton's Second Law
	**Prerequisites**: Understanding of velocity, acceleration, and the effects of constant force.
## SOLUTION PATHWAY
**Given Information**:
	Initial velocity, $u = 0 \, m/s$
	Final velocity after 1 second, $v = 6 \, m/s$
	Time for acceleration, $t_1 = 1 \, s$
	Time for deceleration, $t_2 = 2 \, s$
**Method & Steps**:
	1. **Calculate acceleration:**
		Reasoning: The car experiences constant acceleration due to the electric field.
		Formula/Rule: $v = u + at$
		Working: $6 = 0 + a(1) \implies a = 6 \, m/s^2$
	2. **Calculate displacement in the first second:**
		Working: $s_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + \frac{1}{2}(6)(1)^2 = 3 \, m$
	3. **Calculate the velocity after 2 seconds of deceleration:**
		Reasoning: The reversed field causes deceleration of the same magnitude but opposite direction.
		Working: $v' = v - at_2 = 6 - (6)(2) = -6 \, m/s$
	4. **Calculate displacement during deceleration:**
		Working: $s_2 = vt_2 - \frac{1}{2}at_2^2 = (6)(2) - \frac{1}{2}(6)(2)^2 = 12 - 12 = 0 \, m$
	5. **Calculate total displacement:**
		Working: $s = s_1 + s_2 = 3 + 0 = 3 \, m$
	6. **Calculate average velocity:**
		Formula/Rule: Average velocity = $\frac{	ext{Total Displacement}}{	ext{Total Time}}$
		Working: Average velocity $= \frac{3}{1+2} = \frac{3}{3} = 1 \, m/s$
	7. **Calculate total distance:**
		Reasoning: Distance is the magnitude of displacement for each phase of motion.
		Working: Total distance = $|s_1| + |vt_2 - \frac{1}{2}at_2^2| = 3 + |12-12| = 3 \, m$
	8. **Calculate average speed:**
		Formula/Rule: Average speed = $\frac{	ext{Total Distance}}{	ext{Total Time}}$
		Working: Average speed $= \frac{3}{3} = 1 \, m/s$
	Result: Average velocity = $1 \, m/s$, Average speed = $1 \, m/s$
	Verification: The answer matches option B.
 ## OPTION ANALYSIS
[A] :x: Incorrect. Confuses average speed with average velocity in the case where there is a change in direction.
[B] ✓: Correct. Both average speed and average velocity are calculated correctly.
[C] :x: Incorrect. Average velocity is not zero as there is net displacement.
[D] :x: Incorrect. Average speed is not different from average velocity in this scenario.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing distance and displacement: Displacement considers direction, whereas distance does not. 
		2. Incorrectly applying signs for acceleration and velocity: When the field reverses, acceleration becomes negative.
	**Quick Tips**:
		1. Remember that average velocity is displacement over time, and average speed is distance over time. 
		2. Visualize the motion with a velocity-time graph to understand the changes in velocity and displacement.

Mr. ANIL KUMAR.G · Answer

Average velocity = 2 m/s, Average speed = 2 m/s

Mr. ANIL KUMAR.G · Answer

The correct Option is "Average velocity = 1 m/s, Average speed = 1 m/s"

Mr. ANIL KUMAR.G · Answer

Average velocity = 0 m/s, Average speed = 2 m/s

Mr. ANIL KUMAR.G · Answer

Average velocity = 1 m/s, Average speed = 2 m/s

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