On a $60 k m$ track, a train travels the first $30 k m$ with a uniform speed of $30 k m h^{- 1}$ . How fast must the train travel the next $30 k m$ so as to average $40 k m h^{- 1}$ for the entire trip?

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Solution

Let the speed of the train for next 30 km be "x".
Since, $t = \frac{d}{s}$
So, time taken by the train for first 30 km,
$t_{1} = \frac{30 k m}{30 k m / h} = 1 h$
Time taken by the train to travel next 30 km:
$t_{2} = \frac{30 k m}{x k m / h} = (30 / x) h$
Time taken for the entire trip:
$t = \frac{60 k m}{4 k m / h} = 1.5 h$
So, $1 + \frac{30}{x} = 1.5$
$\Rightarrow x = 60 k m / h$

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On a 60km track, a train travels the first 30km with a uniform speed of 30km h−1. How fast must the train travel the next 30km so as to average 40km h−1 for the entire trip?

Let the speed of the train for next 30 km be "x".Since, t=dsSo, time taken by the train for first 30 km,t1=30km30km/h=1hTime taken by the train to travel next 30 km:t2=30kmxkm/h=(30/x)hTime taken for the entire trip:t=60km4km/h=1.5hSo, 1+30x=1.5⇒x=60km/h

On a $60 k m$ track, a train travels the first $30 k m$ with a uniform speed of $30 k m h^{- 1}$ . How fast must the train travel the next $30 k m$ so as to average $40 k m h^{- 1}$ for the entire trip?