Question

On a $ 60km $ track, a train travels the first $ 30km $ with a speed of $ 30km{h^{ - 1}} $ . How fast must the train travel the next $ 30km $ so as to average $ 40km{h^{ - 1}} $ for the whole trip? Ans- $ 60km{h^{ - 1}} $ A body covers one third of its journey with speed $ 'u' $ , the next one third with speed $ 'y' $ and the last one with $ 3w7w $ .

Answer 1

Hint :First of all find the time taken for the train to travel the first $ 30km $ for which the speed is $ 30km{h^{ - 1}} $ , then let us consider the speed for the next $ 30km $ and find the corresponding time. Now as the average speed for the trip and the whole distance are given, speed for next $ 30km $ can be found.
The formula for the time taken to cover a particular distance
 $ time = \dfrac{{distance}}{{speed}} $

Complete Step By Step Answer:
It is given that the first $ 30km $ is traveled with the speed $ 30km{h^{ - 1}} $ , let us consider that the speed with which the next $ 30km $ is travelled be $ xkm{h^{ - 1}} $
Now as we know that
 $ time = \dfrac{{distance}}{{speed}} $
This implies the time $ {t_1} $ taken to travel the first $ 30km $ will be
 $ {t_1} = \dfrac{{30km}}{{30km{h^{ - 1}}}} = 1h $
The time taken to travel the next $ 30km $ will be
 $ {t_2} = \dfrac{{30km}}{{xkm{h^{ - 1}}}} = \dfrac{{30}}{x}h $
Now as the average speed for the whole trip is $ 40km{h^{ - 1}} $ , so, average time taken for whole trip is
 $ t = \dfrac{{60km}}{{40km{h^{ - 1}}}} = \dfrac{3}{2}h $
This implies
 $ 1h + \dfrac{{30}}{x}h = \dfrac{3}{2}h $
Further solving for the value of $ x $ , we get
$ \Rightarrow \dfrac{{30}}{x} = \dfrac{1}{2} \\
   \Rightarrow x = 60 \\ $
Thus, the train travels the next $ 30km $ with the speed $ 60km{h^{ - 1}} $ which is the same as given in the answer for the question.

Note :
It is important to note the total average time equals the sum of time taken for the first $ 30km $ and the next $ 30km $ . The distance to speed the ratio is a very important factor for solving any distance related questions. You might think of using equations of motion but that is needless as it can be directly solved.