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On a \( 60 \mathrm{~km} \) track, a train travels the first \( 30 \mathrm{~km} \) at a uniform speed of \( 30 \mathrm{~km} / \mathrm{h} \). How fast must the train travel the next \( 30 \mathrm{~km} \) so as to average \( 40 \mathrm{~km} / \mathrm{h} \) for the entire trip?
Here given, Total distance $d_{total}=60\ km$
Average speed $v_{average}=40\ km/h$
Therefore, total time $t_{total}=\frac{d_{total}}{v_{average}}$
$=\frac{60\ km}{40\ km/h}$
Or $t_{total}=1.5\ hr$
For the journey of first $30\ km$:
Distance traveled $d_1=30\ km$
Speed $v_1=30\ km/h$
Therefore, time $t_1=\frac{d_1}{v_1}$
$=\frac{30\ km}{30\ km/h}$
$=1\ hr$
Therefore, time taken for completing the next $30\ km$ distance $t_2=t_{total}-t_1$
$=1.5\ h-1\ h$
Or $t_2=0.5\ h$
Therefore, speed during the next $30\ km$ journey $v_2=\frac{distance(d_2)}{time(t_2)}$
$=\frac{30\ km}{0.5\ h}$
$=60\ km/h$
Thus, the train should travel at the speed of $60\ km/h$ for thee next $30\ km$ so as to average \( 40 \mathrm{~km} / \mathrm{h} \) for the entire trip.