Question

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

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Answer 1

Given :

• Initial velocity, u = 45 km/h = 12.5 m/s
• Final velocity, v = 0 m/s (as it gets stopped)
• Time, t = 6 seconds

To find :

• Retardation
• Distance

According to the question,

By using Newtons first equation of motion,

→ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• t = Time
• a = Acceleration

→ 0 = 12.5 + a × 6

→ 0 - 12.5 = 6a

→ - 12.5 = 6a

→ - 12.5 ÷ 6 = a

→ - 2.08 = a

So,the retardation is - 2.08 m/s².

Now,

→ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

→ (0)² = (12.5)² + 2 × - 2.08 × s

→ 0 = 156.25 + (- 4.16s)

→ 0 - 156.25 = - 4.16s

→ - 156.25 = - 4.16s

→ 156.25 ÷ 4.16 = s

→ 37.56 = s

So,the distance covered by the train is 37.56 m.

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Answer 2

$\huge\bold{\underbrace{Answer}}$

Given data:

• Initial velocity (u) = 45 km/hr (Convert into m/s) = 12.5 m/s
• Final velocity (v)= 0 m/s (As the train applies brake, the body comes to rest)
• Time taken (t) = 6 second

To find :

• Retardation of the train.
• Distance covered by train before coming to rest.

Formula used :

• a = v - u / t
• S = ut + 1/2 at²

Solution :

Firstly let's find out the retardation of train

a = v - u / t

a = 0 - 12.5 / 6

a = - 12.5 / 6

a = -2.083

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Second let's find out the distance covered by the train before coming to rest

s = ut + 1/2 at²

s = 12.5 * 6 + 1/2 (-2.083) 6²

s = 75+ 1/2 (-2.083) 36

s = 75 + 1/2 (-74.988)

s = 75 + (-37.494)

s = 75 - 37.494

s = 37.506

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Hence,

• Retardation of the train is -2.083 m/s
• distance covered by the train before coming to rest is 37.506 m