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
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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