A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE , its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the 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 ?

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE , its velocity increases from 0 to 6 m/s in one second duration. At that  instant the direction of the 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 ?

(a) 2 m/s, 4 m/s

(b) 1 m/s, 3 m/s

(c) 1.5 m/s, 3 m/s

(d) 1 m/s, 3.5 m/s

Answer :  (b) 1 m/s, 3 m/s

Initial velocity: \( u = 0 \)
Final velocity: \( v = 6\, \text{m/s} \)
Time: \( t = 1\, \text{s} \)

Acceleration:

\( a = \frac{v - u}{t} \) \( = \frac{6 - 0}{1} = 6\, \text{m/s}^2 \)

Displacement:

\( s = ut + \frac{1}{2} a t^2 \) \( = 0 + \frac{1}{2} \cdot 6 \cdot 1^2 = 3\, \text{m} \)

Initial velocity: \( u = 6\, \text{m/s} \)
Acceleration: \( a = -6\, \text{m/s}^2 \)

Final velocity:

\( v = u + at \) \( = 6 + (-6)(1) = 0\, \text{m/s} \)

Displacement:

\( s = ut + \frac{1}{2} a t^2 \) \( = 6(1) + \frac{1}{2}(-6)(1)^2 = 6 - 3 = 3\, \text{m} \)

\( u = 0\, \text{m/s} \)

\( a = -6\, \text{m/s}^2 \)

Final velocity:

\( v = u + at \) \( = 0 + (-6)(1) = -6\, \text{m/s} \)

Displacement:

\( s = ut + \frac{1}{2} a t^2 \) \( = 0 + \frac{1}{2}(-6)(1)^2 = -3\, \text{m} \)

Average Velocity:

\( \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} \)

\( = \frac{3 + 3 + (-3)}{1 + 1 + 1} = \frac{3 \, \text{m}}{3 \, \text{s}} = 1 \, \text{m/s} \)

Average Speed:

\( \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \)

\( = \frac{|3| + |3| + |-3|}{1 + 1 + 1} = \frac{9 \, \text{m}}{3 \, \text{s}} = 3 \, \text{m/s} \)