A train is targeted from Delhi to Mumbai at an average speed of 80 km/h but due to repairs of track loses 2 hrs in the first part of the journey. If then accelerates at a rate of 20 km/h2 till the speed reaches 100 km/h. Its speed now maintained till the end of the journey. If the train now reaches station in time. Find the distance from when it started accelerating.
Answers
Answer:
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Explanation:
If then accelerates at a rate of 20 kph^(2) till the speed reaches 100 kph. Its speed is now maintained till the end ofthe journey. ... to Pune at an average speed of 80 kph but due to repairs of track looses 2hr sin the first part of the journey. If ... A carrier train, when it is 100 km away from the station, going at.
Given : A train is targeted from Delhi to Mumbai at an average speed of 80 km/h but due to repairs of track loses 2 hrs in the first part of the journey. Then accelerates at a rate of 20 km/hr² till the speed reaches 100 km/h. Its speed now maintained till the end of the journey. The train now reaches station in time
To find : distance from when it started accelerating.
Solution:
Average speed = 80 km/hr
Total time taken = T hr
Distance = 80T km
loses 2 hrs in the first part of the journey
Distance covered = 80 (x - 2) km till start accelerating
Remaining hrs = T - x hrs
accelerates at a rate of 20 km/hr till the speed reaches 100 km/h
100 = 80 + (20)T => T = 1 hr
Distance covered in 1 hr = 1 * (80 + 100)/2 = 90 km
Remaining time = T - x - 1 hr
Distance covered = 100(T - x - 1) km
80(x - 2) + 90 + 100(T - x - 1) = 80T
=> 80x - 160 + 90 + 100T - 100x - 100 = 80T
=> 20T - 20x = 170
=> T - x = 8.5
=> x = T - 8.5
80 (x - 2) = 80 (T - 8.5 - 2) = 80T - 840 km
840 km before reaching Mumbai
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