A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s. It then pro

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A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s. It then proceeds at constant speed for 1100m before slowing down at 2.2m/s^2 until it stops at the station. A) What is the distance between stations? B) How much time does it take the train to go between the stations?

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Orla Orla 6 months 2021-07-30T04:19:54+00:00 1 Answers 11 views 0

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    2021-07-30T04:21:21+00:00

    Answer:

    A) The distance between the stations is 1430m

    B) The time it takes the train to go between the stations is 80s

    Explanation:

    First we will calculate the distance covered for the first 20s.

    From one the equations of kinematics for linear motion

    S = ut + \frac{1}{2}at^{2}  \\

    Where S is distance traveled

    u is the initial velocity

    t is time

    and a is acceleration

    Since the train starts from rest, u = 0 m/s

    Hence,  for the first 20s

    a = 1.1 m/s²; t = 20s,  u = 0 m/s

    S = ut + \frac{1}{2}at^{2}  \\ gives

    S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}

    S =  \frac{1}{2}(1.1)(20)^{2}

    S = 220m

    This is the distance covered in the first 20s.

    • The train then proceeds at constant speed for 1100m.

    Now, we will calculate the speed attained here

    From

    v = u +at

    Where v is the final velocity

    Hence,

    v = 0 + 1.1(20)

    v = 1.1(20)

    v = 22 m/s

    This is the constant speed attained when it proceeds for 1100m

    • The train then slows down at a rate of 2.2 m/s² until it stops

    We can calculate the distance covered while slowing down from

    v^{2} = u^{2} + 2as

    The initial velocity, u here will be the final velocity before it started slowing down

    u = 22 m/s

    The final velocity will be 0, since it came to a stop.

    v = 0 m/s

    a = -2.2 m/s² ( – indicates deceleration)

    Hence,

    v^{2} = u^{2} + 2as gives

    0^{2} =22^{2} +2(-2.2)s

    0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m

    This is the distance traveled while slowing down.

    A) The distance between the stations is

    220m + 1100m + 110m

    = 1430m

    Hence, the distance between the stations is 1430m

    B) The time it takes the train to go between the stations

    The time spent while accelerating at 1.1 m/s² is 20s

    We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,

    From,

    Speed =\frac{Distance}{Time}\\

    Then,

    Time = \frac{Distance}{Speed}

    Time = \frac{1100}{22}

    Time = 50 s

    And then, the time spent while decelerating (that is, while slowing down)

    From,

    v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s

    This is the time spent while slowing down until it stops at the station.

    Hence, The time it takes the train to go between the stations is

    20s + 50s + 10s = 80s

    The time it takes the train to go between the stations is 80s

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