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(B) The acceleration $a$ is given by $a = \frac{v - u}{t} = \frac{6 - 0}{1} = 6\, m s^{-2}$.For the interval $t = 0$ to $t = 1\, s$:The displacement $

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$1\, m/s, 3\, m/s$

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(B) The acceleration $a$ is given by $a = \frac{v - u}{t} = \frac{6 - 0}{1} = 6\, m s^{-2}$.For the interval $t = 0$ to $t = 1\, s$:The displacement $S_1 = u t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} 	imes 6 	imes (1)^2 = 3\, m$.At $t = 1\, s$,the field is reversed,so the new acceleration is $a' = -6\, m s^{-2}$.For the interval $t = 1\, s$ to $t = 2\, s$:The displacement $S_2 = v_1 t + \frac{1}{2} a' t^2 = 6 	imes 1 + \frac{1}{2} 	imes (-6) 	imes (1)^2 = 6 - 3 = 3\, m$.For the interval $t = 2\, s$ to $t = 3\, s$:The velocity at $t = 2\, s$ is $v_2 = v_1 + a' t = 6 + (-6) 	imes 1 = 0\, m s^{-1}$.The displacement $S_3 = v_2 t + \frac{1}{2} a' t^2 = 0 	imes 1 + \frac{1}{2} 	imes (-6) 	imes (1)^2 = -3\, m$.Total displacement $S = S_1 + S_2 + S_3 = 3 + 3 - 3 = 3\, m$.Average velocity = $\frac{	ext{Total displacement}}{	ext{Total time}} = \frac{3}{3} = 1\, m s^{-1}$.Total distance = $|S_1| + |S_2| + |S_3| = 3 + 3 + 3 = 9\, m$.Average speed = $\frac{	ext{Total distance}}{	ext{Total time}} = \frac{9}{3} = 3\, m s^{-1}$.

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