A train covers a distance of $200 k m$ daily from its origin to destination in $5 h o u r s$ . On a particular day, the train covers the distance of $160 k m$ in $4.5 h o u r s$ . What should be the speed of the train after $4.5 h o u r s$ such that it reaches the destination on time?

50 k m h^{- 1}

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80 k m h^{- 1}

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100 k m h^{- 1}

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70 k m h^{- 1}

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Solution

The correct option is B $80 k m h^{- 1}$
Given:
Distance covered in the first part of the journey, $d_{1} = 160 k m$
Time taked for the first part of the journey, $t_{1} = 4.5 k m$
Total distance to covered by the train, $d = 200 k m$
Total time to be taken, $t = 5 h$

Remaining distance to be covered:
$d_{2} = d - d_{1}$
$d_{2} = 200 - 160 = 40 k m$
Remaining time available:
$t_{2} = t - t_{1}$
$t_{2} = 5 - 4.5 = 0.5 h$
Speed is defined as the distance covered per unit time.
$S p e e d = \frac{D i s t a n c e}{T i m e}$
Thus, required speed:
$v = \frac{d_{2}}{t_{2}}$
$v = \frac{40 k m}{0.5 h} = 80 k m h^{- 1}$

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A train covers a distance of 200 km daily from its origin to destination in 5 hours. On a particular day, the train covers the distance of 160 km in 4.5 hours. What should be the speed of the train after 4.5 hours such that it reaches the destination on time?

A train covers a distance of $200 k m$ daily from its origin to destination in $5 h o u r s$ . On a particular day, the train covers the distance of $160 k m$ in $4.5 h o u r s$ . What should be the speed of the train after $4.5 h o u r s$ such that it reaches the destination on time?