A train 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?

Question

Jamie Anthony

Physics

28 November, 17:42

A train 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?

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Answers (1)

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Georgia Fowler · Answer 1

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