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Physics Question on Electric charges and fields

A toy car with charge qq moves on a frictionless horizontal plane surface under the influence of a uniform electric field EE . Due to the force q E⃗q\,\vec{E} , its velocity increases from 0 to 6m/s−10 \,to\, 6 m/s^{-1} 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 30 \,to\, 3 seconds are respectively

A

1.5 ms−1,3ms−11.5\, ms^{-1} , 3 ms^{-1}

B

2ms−1,4ms−12 ms^{-1} , 4 ms^{-1}

C

1ms−1,3.5ms−11 ms^{-1} , 3.5 ms^{-1}

D

1ms−1,3ms−11 ms^{-1} , 3 ms^{-1}

Answer

1ms−1,3ms−11 ms^{-1} , 3 ms^{-1}

Explanation

Solution

Acceleration a =6−01=6 ms−2=\frac{6-0}{1}=6\, ms ^{-2}
For t=0t=0 to t=1 st=1 \,s
S1=12×6(1)2=3 m…S_{1}=\frac{1}{2} \times 6(1)^{2}=3\, m \ldots...(i)
For t=1 st=1 \,s to t=2 st=2 \,s,
S2=6.1−12×6(1)2=3 m….S_{2}=6.1-\frac{1}{2} \times 6(1)^{2}=3 \,m \ldots . (ii)
For t=2 st=2\,s to t=3 st=3 \,s
S3=0−12×6(1)2=−3 mS_{3}=0-\frac{1}{2} \times 6(1)^{2}=-3 \,m....(iii)
Total displacement S=S1+S2+S3=3 mS=S_{1}+S_{2}+S_{3}=3\, m
Average velocity =33=1 ms−1=\frac{3}{3}=1\, ms ^{-1}
Total distance travelled =9 m=9 \,m
Average speed =93=3 ms−1=\frac{9}{3}=3\, m s^{-1}