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A train travels along a straight horizontal track with constant acceleration. Points A,...

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saj-94 | (Level 1) Salutatorian

Posted August 6, 2013 at 8:53 AM via web

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A train travels along a straight horizontal track with constant acceleration. Points A, B and C are on the track with B between A and C. The distance AB is 1200 m and the distance BC is 2500 m. As the train passes B, its speed is 26ms–1. The train takes 60 s to travel from A to B.

c) Calculate the speed of the train as it passes C, giving your answer correct to one decimal place.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 6, 2013 at 9:25 AM (Answer #1)

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The previous parts of this question are answered in;

http://www.enotes.com/homework-help/train-travels-along-straight-horizontal-track-446431

http://www.enotes.com/homework-help/train-travels-along-straight-horizontal-track-446432

It is given that the train travels 2500m from B to C.

Acceleration of the train is `0.2m/s^2` .

See the attached image with the calculations.

Let us say the velocity of C is Um/s and the time taken to travel between B and C as T seconds.


The acceleration from B to C is 0.2m/s^2.

`(U-26)/T = 0.2`

`U = 26+0.2T`

The area of the time velocity graph from B to C represents the distance travelled from B to C.

`2500 = (U+26)/2xxT`

`2500 = (26+0.2T+26)/2xxT`

`5000 = 52T+0.2T^2`

`0.2T^2+52T-5000 = 0`

`T = (-52+-sqrt((-52)^2-4xx0.2xx(-5000)))/(2xx0.2)`

`T = 74.7 `

`Or `

`T = -334.7`


`T = 74.7` since it is greater than 0.

`U = 26+0.2T = 40.9`

So the speed of the train at C is 40.9m/s.

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