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(A) Let the acceleration be $\alpha = 0.2 \, m/s^2$,retardation be $\beta = 0.4 \, m/s^2$,and total time be $T = 30 	imes 60 = 1800 \, s$.Let $t_1$ b

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$216$

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(A) Let the acceleration be $\alpha = 0.2 \, m/s^2$,retardation be $\beta = 0.4 \, m/s^2$,and total time be $T = 30 	imes 60 = 1800 \, s$.Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation. Then $t_1 + t_2 = T$.The maximum velocity reached is $v = \alpha t_1 = \beta t_2$.Thus,$t_1 = \frac{v}{\alpha}$ and $t_2 = \frac{v}{\beta}$.Substituting into the time equation: $v(\frac{1}{\alpha} + \frac{1}{\beta}) = T \Rightarrow v = \frac{\alpha \beta T}{\alpha + \beta}$.The total distance $S$ is the area under the velocity-time graph,which is a triangle: $S = \frac{1}{2} 	imes v 	imes T$.Substituting $v$: $S = \frac{1}{2} 	imes \frac{\alpha \beta T^2}{\alpha + \beta}$.Given $\alpha = 0.2 \, m/s^2$,$\beta = 0.4 \, m/s^2$,and $T = 1800 \, s$:$S = \frac{1}{2} 	imes \frac{0.2 	imes 0.4}{0.2 + 0.4} 	imes (1800)^2$.$S = \frac{1}{2} 	imes \frac{0.08}{0.6} 	imes 3240000 = \frac{1}{2} 	imes \frac{2}{15} 	imes 3240000 = 216000 \, m$.$S = 216 \, km$.

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