Atrain accelerates from a station with a = 1.841m/s ? upon reaching a speed of 23.52m/s the train travels at a constant velocity for a period. the train slows down as it approaches the next station at a rate of 2m/s ? and stops at that station hint: sketch a graph of uversus tfor the train's journey. (a) of the two stations are 1,200m apart, how long does the train journey take?

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 obtaint = 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 × fwhere 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

Thenwe 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

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