The maximum possible acceleration of a train moving on a straight track is 10 m / s 2 and maximum possible retardation is 5 m / s 2 . If maximum achievable speed of train is 10 m / s then minimum time

Question

The maximum possible acceleration of a train moving on a straight track is $10 m / s^{2}$ and maximum possible retardation is $5 m / s^{2}$ . If maximum achievable speed of train is $10 m / s$ then minimum time in which train can complete a journey of 135 m starting from rest and ending at rest, is :-

Options:

Answer:

C

Solution:

$\frac{1}{2} (1 + 2) (10) + 10 t = 135$

$t = 12 s$

∴ (C)

TestHub · Accepted Answer

2 1 ​ ( 1 + 2 ) ( 10 ) + 10 t = 135 t = 12 s ∴ (C)

Physics - Kinematics Question with Solution | TestHub

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