A train moving on linear path travles a distance D at a constant speed of 30km/h , then it travles the same distance in opposite direction and reaches initial position at a constant speed of 45km/h. what is the average speed of the train?
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As the initial speed to cover distance D is 30 km/h, and later it increases to 45 km/h, we can conclude that the time taken by return journey would have been 2/3 of the journey upwards.
i.e. if it took x hours to travel distance D initially at the speed of 30 km/hr, it would need 2*x/3 hours at the speed of 45 km/h.
The average speed would hence be,
= Total distance traveled / total time taken
= 2*D / (x + (2*x/3))
= 6*D/ 5*x
= (6/5) *(D/x)
= (6/5) * (30) { as the distance D is covered in x hour at the speed of 30 km/h)
= 36 km/h
The average speed = ((2*v1*v2)/(v1+v2))
= (2 * 30 * 45) / (30 + 45)
= 2700 / 75
= 36 km/h
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