Question

Assume the rate of consumption of coal by a locomotive varies as the square of the speed and is 1000 kg per hour when the speed is 60 km per hour. if the coal costs the railway company ₹15 per 100 kg and if the other expenses of the train be ₹12 per hour, find the formula for the cost in paise per kilometer when the speed is s km per hour.

Answer 1

Step-by-step explanation:

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Answer 2

The formula for the cost in paise per kilometer when the speed is $s$ km per hour is $\frac{1200}{s}+\frac{25s}{6}$

Step-by-step explanation:

Given:Rate of consumption of coal by a locomotive varies as the square of the speed$= R=KS^2$, Rate of consumption of coal when the speed is $60$ km per hour$=1000$ Kg per hour, Cost of coal to the railway company$=$Rs$15$ per $100$Kg, Other expenses of the train be Rs$12$ per hour

To Find: The formula for the cost in paise per Kilometer when the speed is s km per hour

Concept/Formula used:

$R=KS^2$

Here,$K$ is an arbitrary constant

When $R=1000$kg per hr; $S=60$Kmph

$1000=K60^2$

$K=\frac{5}{18}$

Hence,$R=\frac{5}{18} s^2$

Cost of coal to the railway company$=\frac{15}{100}$ Rs

Cost of R $= \frac{5}{18} s^2\frac{15}{100}$ Rs$=\frac{5}{18}s^215$ paise

Cost of coal for $s$ km$=\frac{5}{18} s^215$

Cost of coal per km$=\frac{5}{18}s^2\frac{15}{s}$

Other expenses of train per km$=\frac{1200}{s}$paise

Therefore,total cost in paise per km

$= \frac{1200}{s} +\frac{5}{18} s^2\frac{15}{s}$

$=\frac{1200}{s} +\frac{25s}{6}$

The formula for the cost in paise per kilometer when the speed is $s$ km per hour is$\frac{1200}{s} +\frac{25s}{6}$