Answer:
63.29s
Explanation:
Firstly calculate the time taken to reach 23.52m/s
;use the formula...v = u + at
23.52 = 0 + 1.841t
then obtain...t = 12.78s
Then calculate the time for the last part of the journey...where the train slows down...
use the formula that is above...
0 = 23.52 - 2t...(negative for deceleration)
then obtain....t = 11.76s
Then we know that the total area under the graph of u against t..is equal to 1200m
For the first triangle(first part of the journey...where the train accelerates)
(23.52 × 12.78)÷2 = 150.3m
Then for the constant velocity part...a rectangle...
23.52 × f.....where f represents the time taken by the train having constant velocity.
...= 23.52fm
Then for the last part of the journey...the deceleration part..a triangle
(23.52 × 11.76)÷ 2 = 138.3m
Then....we add all the obtained distances and equate to 1200m so that we can obtain time (f)
138.3 + 150.3 + 23.52f = 1200
where f = 38.75s
Then total time for the whole journey of the train...
38.75 + 11.76 + 12.78
;Ans = 63.29s
