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    Question:
    A Student, standing on the platform at a railway station, notices that the first two carriages on an arriving train pass her in 2.0 seconds, and the the next two in 2.4 seconds. The train is decelerating uniformly. Each carriage is 20m long. When the train stops, the student is opposite the last carriage. How many carriages are there in the train?

    My Calculations:
    First I began calculating the velocity of the incoming carriages:
    40m/2 seconds = 20 m/s
    40m/2.4 second = 16.6 m/s

    Then using v = u + at. I re-arrange it to get the acceleration.
    (16.6 - 20) / 4.4 = - 0.772 m / s^2

    Then I find the total time needed until the train reaches a stop from an initial velocity of 16.6 m/s .
    so ... (0 - 16.6) / -0.772 = 21.5 seconds

    Since acceleration is uniform I then use: s = ut + 1/2 at^2 to find the distance of part of the train.
    s = 16.6 * 21.5 + 1/2 * -0.772 * 21.5^2. This gives 178 m. Round it to 180. Divide this by 20 as a carriage is 20m long. This gives 9 carriages + 4 from the start so 13 carriages overall. However at the back of the book it says the answer is 8. Can someone tell me where I went wrong or show the proper method of working it out?
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    (Original post by zattyzatzat)
    Question:
    A Student, standing on the platform at a railway station, notices that the first two carriages on an arriving train pass her in 2.0 seconds, and the the next two in 2.4 seconds. The train is decelerating uniformly. Each carriage is 20m long. When the train stops, the student is opposite the last carriage. How many carriages are there in the train?

    My Calculations:
    First I began calculating the velocity of the incoming carriages:
    40m/2 seconds = 20 m/s
    40m/2.4 second = 16.6 m/s

    Then using v = u + at. I re-arrange it to get the acceleration.
    (16.6 - 20) / 4.4 = - 0.772 m / s^2

    Then I find the total time needed until the train reaches a stop from an initial velocity of 16.6 m/s .
    so ... (0 - 16.6) / -0.772 = 21.5 seconds

    Since acceleration is uniform I then use: s = ut + 1/2 at^2 to find the distance of part of the train.
    s = 16.6 * 21.5 + 1/2 * -0.772 * 21.5^2. This gives 178 m. Round it to 180. Divide this by 20 as a carriage is 20m long. This gives 9 carriages + 4 from the start so 13 carriages overall. However at the back of the book it says the answer is 8. Can someone tell me where I went wrong or show the proper method of working it out?
    never mind accidentally took the total time of 4.4 instead of 2.4. Found my error
 
 
 
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