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?

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

it should be -8.6 cm hope this out

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explanation:

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