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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 a 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 m/s, 3.5 m/s

D

1.5 m/s, 3 m/s

Step-by-Step Solution

  1. 0 to 1 sec: Initial velocity u=0u=0, Final velocity v=6v=6 m/s. Acceleration a=(6−0)/1=6a = (6-0)/1 = 6 m/s². Displacement S1=12at2=12(6)(1)2=3S_1 = \frac{1}{2}at^2 = \frac{1}{2}(6)(1)^2 = 3 m. Distance d1=3d_1 = 3 m.
  2. 1 to 3 sec: Field reverses, so acceleration a′=−6a' = -6 m/s². Initial velocity for this interval u′=6u' = 6 m/s. Duration t′=2t' = 2 s.
  • Displacement S2=u′t′+12a′(t′)2=6(2)+12(−6)(2)2=12−12=0S_2 = u't' + \frac{1}{2}a'(t')^2 = 6(2) + \frac{1}{2}(-6)(2)^2 = 12 - 12 = 0 m.
  • To find distance, check if particle stops. v=u′+a′t⇒0=6−6t⇒t=1v = u' + a't \Rightarrow 0 = 6 - 6t \Rightarrow t=1 s. So it stops at total time t=2t=2 s and reverses.
  • Distance forward (1s to 2s): dfwd=6(1)−3(1)2=3d_{fwd} = 6(1) - 3(1)^2 = 3 m.
  • Distance backward (2s to 3s): Magnitude of displacement for remaining 1s = 3 m.
  • Total distance d2=3+3=6d_2 = 3 + 3 = 6 m.
  1. Averages (0 to 3 sec):
  • Total Displacement = S1+S2=3+0=3S_1 + S_2 = 3 + 0 = 3 m.
  • Total Distance = d1+d2=3+6=9d_1 + d_2 = 3 + 6 = 9 m.
  • Average Velocity = 3 m/3 s=13 \text{ m} / 3 \text{ s} = 1 m/s.
  • Average Speed = 9 m/3 s=39 \text{ m} / 3 \text{ s} = 3 m/s.
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