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22 January, 05:35

How many seconds does it take to reach 18 m/s starting from rest if the train's mass, including its load, is 5.15 * 104 kg, assuming 95.0% efficiency and constant power?

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  1. 22 January, 09:16
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    a) Power consumption rate = 409.5 KW

    b) Time taken to reach 18 m/s = 42.86 s

    Explanation:

    a) Power is given by P = IV

    where I = current drawn = 630 A

    V = voltage = 650 V

    P = 630 * 650 = 409500 W = 409.5 KW

    b) Power is also given as P = Force * velocity

    Force = ?

    v = 18 m/s

    Power = 95% of 409500 = 389025 W (95% efficiency)

    Force = Power/velocity = 389025/18 = 21612.5 N

    Force = mass * acceleration = 5.15 * 10⁴ * a

    51500a = 21612.5

    a = 21612.5/51500 = 0.42 m/s²

    u = initial velocity = 0 m/s

    v = final velocity = 18 m/s

    a = 0.42 m/s²

    t = ?

    v = u + at

    18 = 0 + 0.42t

    t = 18/0.42 = 42.86 s
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