# 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.

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Updated on: 10-Oct-2022

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