Question

.A train was moving at the rate of 54Kmh-1 When brakes were applied. It came to rest within a
distance of 225m.Calculate the retardation produced in the train

Answer 1

Answer:

Explanation:

Given,

Initial velocity of the train, u = 54 km/h = 54 × 5/18 = 15 m/s

Final velocity of the train, v = 0 m/s (As brakes applied)

Distance covered by the train, s = 225 m.

To Find,

Retardation by the train, a = ?

Formula to be used,

3rd equation of motion,

i.e, v² - u² = 2as

Solution,

Putting all the values, we get

v² - u² = 2as

⇒ (0)² - (15)² = 2 × a × 225

⇒ 225 = 450a

⇒ 225/450 = a

⇒ a = - 0.5 m/s²

Hence, the retardation pruduced by train is - 0.5 m/s².

Answer 2

Answer:

Given :-

• Initial velocity of train = 54 km/h × 5/18 = 15 m/s
• Final velocity of train = 0 m/s
• Distance travelled by train = 225 m

To Find :-

Retardation

SoluTion :-

Here we will use newton third Equation

$\huge \bf \: {v}^{2} - {u}^{2} = 2as$

$\sf\: {0}^{2} - {15}^{2} = 2(a)(225)$

$\sf \: 0 - 225 = 450(a)$

$\sf \: - 225 = 450a$

$\sf \: \dfrac{ - 225}{450} = a$

$\sf \: a = - 0.5 \: mps {}^{2}$

Hence :-

The retardation of train is -0.5 m/s².