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On a `80 km` track, a train travels `40 km` with a uniform speed of `30 km h^(-1)`. How fast must the train travel the next `40 km h^(-1)` as to have average speed `40 km h^(-1)` for the entire trip? |
Answer» Here, `s_(1) =40 km, v_(1) =30 km h^(-1) s_(2) =40 km, v_(av) =40 km h^(-1), v_(2)=?` `v_(av)=(S_(1)+S_(2))/(S_(1)/v_(1) +S_(2)/v_(2)) =((40+40)/((40)/(30) +(40)/v_(2)) =(2xx 30 v_(2))/(30 +v_(2))` or `40 =(60 v_(2) )/(30 + v_(2)) or v_(2) =60 km h^(-1)`. |
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