Official (Closed) – Non Sensitive MEF Tutorial 2 Q3 A train with a maximum speed of 29.17 m/s has an acceleration rate of 0.2

Question

Official (Closed) – Non Sensitive
MEF Tutorial 2 Q3
A train with a maximum speed of 29.17 m/s has an
acceleration rate of 0.25 m/s2 and a deceleration
rate of 0.7 m/s2. Determine the minimum running
time, if it starts from rest at one station and stops
at the next station 7 km away.​

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Huyền Thanh 3 years 2021-09-05T07:41:38+00:00 1 Answers 23 views 0

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  1. yenoanh
    0
    2021-09-05T07:43:37+00:00
    Reply

    Answer:

    The minimum running time is 319.47 s.

    Explanation:

    First we find the distance covered and time taken by the train to reach its maximum speed:

    We have:

    Initial Speed = Vi = 0 m/s    (Since, train is initially at rest)

    Final Speed = Vf = 29.17 m/s

    Acceleration = a = 0.25 m/s²

    Distance Covered to reach maximum speed = s₁

    Time taken to reach maximum speed = t₁

    Using 1st equation of motion:

    Vf = Vi + at₁

    t₁ = (Vf – Vi)/a

    t₁ = (29.17 m/s – 0 m/s)/(0.25 m/s²)

    t₁ = 116.68 s

    Using 2nd equation of motion:

    s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

    s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

    s₁ = 1701.78 m = 1.7 km

    Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

    We have:

    Final Speed = Vf = 0 m/s    (Since, train is finally stops)

    Initial Speed = Vi = 29.17 m/s     (The train must maintain max. speed for min time)

    Deceleration = a = – 0.7 m/s²

    Distance Covered to stop = s₂

    Time taken to stop = t₂

    Using 1st equation of motion:

    Vf = Vi + at₂

    t₂ = (Vf – Vi)/a

    t₂ = (0 m/s – 29.17 m/s)/(- 0.7 m/s²)

    t₂ = 41.67 s

    Using 2nd equation of motion:

    s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

    s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

    s₂ = 607.78 m = 0.6 km

    Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

    The remaining distance is:

    s₃ = 7 km – s₂ – s₁

    s₃ = 7 km – 0.6 km – 1.7 km

    s₃ = 4.7 km

    Now, for uniform speed we use the relation:

    s₃ = vt₃

    t₃ = s₃/v

    t₃ = (4700 m)/(29.17 m/s)

    t₃ = 161.12 s

    So, the minimum running time will be:

    t = t₁ + t₂ + t₃

    t = 116.68 s + 41.67 s + 161.12 s

    t = 319.47 s

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