On a 120 km track a train travels the first 30 km at a uniform speed of 30 km per hour.how fast must the train in the next 90km so as to average 60km/hour for the entire trip?
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Hi. ..
distance d1 = 30 km, v1 = 30 kmph
=> time taken = t1 = d1/v1 = 1 hour
So distance remaining = d2 = 120 - d1 = 90 km
let speed to cover distance d2 = v2 kmph
=> time taken = d2/v2 = 90/v2 hours
average speed = total distance / total duration of time.
60 kmph = 120 km / [1 hour + 90 / v2 hours]
=> 1 = 2 / [ 1 + 90 / v2 ]
=> v2 = 90 kmph
====================================
another way:
as the total distance = 120 km
average speed = 60 kmph
=> total time duration = 120 / 60 = 2 hours
time taken for the first 30 km = 30 km /30 kmph = 1 hours
Hence, time remaining to cover the remaining 90 km = 2 hrs - 1 hrs = 1 hour
=> speed during 90 km = 90 kmph..
Hope this helps u!!
distance d1 = 30 km, v1 = 30 kmph
=> time taken = t1 = d1/v1 = 1 hour
So distance remaining = d2 = 120 - d1 = 90 km
let speed to cover distance d2 = v2 kmph
=> time taken = d2/v2 = 90/v2 hours
average speed = total distance / total duration of time.
60 kmph = 120 km / [1 hour + 90 / v2 hours]
=> 1 = 2 / [ 1 + 90 / v2 ]
=> v2 = 90 kmph
====================================
another way:
as the total distance = 120 km
average speed = 60 kmph
=> total time duration = 120 / 60 = 2 hours
time taken for the first 30 km = 30 km /30 kmph = 1 hours
Hence, time remaining to cover the remaining 90 km = 2 hrs - 1 hrs = 1 hour
=> speed during 90 km = 90 kmph..
Hope this helps u!!
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