Question

A train is moving with a velocity of 45 km/hr. The driver applies the brake and the train gets stopped in 6 seconds. Now, calculate retardation as well as distance covered by the train before coming to the rest.

Answer 1

GiveN :

• Initial velocity (u) = 45 km/h = 12.5 m/s
• Final velocity (v) = 0 m/s
• Time interval (t) = 6 s

To FinD :

• Retardation of train
• Distance travelled by train

SolutioN :

Use 1st equation of motion :

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$\implies$ v = u + at

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$\implies$ 0 = -12.5 + 6*a

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$\implies$ 6a = -12.5

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$\implies$ a = 12.5/6

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$\implies$ a = -2.08

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$\implies$ a = -2 m/s² (approx.)

_________________________

Now, use 3rd equation of motion :

$\implies$ v² - u² = 2as

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$\implies$ 0² - 12.5² = -2*2*s

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$\implies$ 4s = 156.25

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$\implies$ s = 156.25/4

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$\implies$ s = 39.06 m (approx.)

Answer 2

GiveN :

Initial velocity (u) = 45 km/h = 12.5 m/s

Final velocity (v) = 0 m/s

Time interval (t) = 6 s

To FinD :

Retardation of train

Distance travelled by train

SolutioN :

Use 1st equation of motion :

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⟹ v = u + at

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⟹ 0 = -12.5 + 6*a

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⟹ 6a = -12.5

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⟹ a = 12.5/6

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⟹ a = -2.08

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⟹ a = -2 m/s² (approx.)

_________________________

Now, use 3rd equation of motion :

⟹ v² - u² = 2as

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⟹ 0² - 12.5² = -2*2*s

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⟹ 4s = 156.25

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⟹ s = 156.25/4

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⟹ s = 39.06 m (approx.)