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NEET Hard

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E⃗\vec{E}. Due to the force qE⃗q\vec{E}, 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

1

1 m/s, 3.5 m/s

2

1 m/s, 3 m/s

3

2 m/s, 4 m/s

4

1.5 m/s, 3 m/s

Step-by-Step Solution

Acceleration a=v/t=6/1=6 m/s2a = v/t = 6/1 = 6 \text{ m/s}^2. At t=1t=1s, v=6v=6 m/s. For t=1t=1 to 33s, a′=−6 m/s2a' = -6 \text{ m/s}^2. Velocity at t=3t=3s is v3=6+(−6)(2)=−6v_3 = 6 + (-6)(2) = -6 m/s. Displacement s=s1+s2=12(6)(1)+[6(2)+12(−6)(22)]=3+[12−12]=3s = s_1 + s_2 = \frac{1}{2}(6)(1) + [6(2) + \frac{1}{2}(-6)(2^2)] = 3 + [12 - 12] = 3 m. Average velocity = 3/3=13/3 = 1 m/s. Distance d=d1+d2=3+∣12−12∣=3+6=9d = d_1 + d_2 = 3 + |12 - 12| = 3 + 6 = 9 m. Average speed = 9/3=39/3 = 3 m/s.

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