On a 120 km track,a train travels the first 30 km at a uniform speed of 30 km/h. How fast must the train travel the next 90 km so as to attain average speed of 60 km/h for the entire trip?

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Solution

Dear Student ,
$S p e e d o f t h e t r a i n i n t h e f i r s t 30 k m, v_{1} = 30 k m / h S p e e d o f t h e t r a i n i n t h e n e x t 90 k m v_{2} (t o b e d e t e r m i n e d) A v e r a g e s p e e d = 60 k m / h \frac{(v_{1} + v_{2})}{2} = 60 k m / h \frac{(30 k m / h + v_{2})}{2} = 60 k m / h \Rightarrow v_{2} = 90 k m / h$

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On a 120 km long track, a train travels the first 30 km with a uniform speed of 30 km/h. How fast must the train travel for the rest part of the journey so as to average 60 km/h for the entire trip?

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On a 120 km track,a train travels the first 30 km at a uniform speed of 30 km/h. How fast must the train travel the next 90 km so as to attain average speed of 60 km/h for the entire trip?

Dear Student , Speed of the train in the first 30 km, v1 = 30 km/h Speed of the train in the next 90 km v2 (to be determined) Average speed = 60 km/h (v1 + v2)2 = 60 km/h(30 km/h + v2)2 =60 km/h ⇒ v2 = 90 km/h Regards