Question

A train running at 108 km/h is brought to a halt in 2 minutes, calculate the retardation produced by the application of breaks. Also calculate the distance the train travels before stopping​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The term 'retardation' is defined as the negative acceleration caused by an object. The formula to calculate retardation is given as:

$\implies \boxed{ \textbf{Retardation } = \bf{- \dfrac{(v-u)}{t}}}$

According to the question,

Initial Velocity = 108 km/hr

Converting it to m/s, we get:

$\implies \text{Initial Velocity (in m/s)} = \dfrac{5}{18} \times 108\\\\\\\implies \boxed{\textbf{Initial Velocity (in m/s)} = \bf{30}}$

Since the train is stopping after 2 minutes, the Final Velocity would be 0 m/s. Converting time to seconds we get:

⇒ 2 min = 2 × 60 = 120 seconds

Substituting the values in the formula we get:

$\implies \text{Retardation} = - \dfrac{ 30 - 0}{120}\\\\\\\implies \text{Retardation} = - \dfrac{30}{120}\\\\\\\implies \boxed{ \textbf{Retardation} = \bf{- 0.25 \:\:m/s^2 } }$

Hence the retardation of the train is - 0.25 m/s².

Distance after which the train stops is given by the second equation of the motion, which is:

⇒ v² - u² = 2as

⇒ 0² - (30)² = 2 ( - 0.25 ) ( s )

⇒ s = - (30)² / - 0.5

⇒ s = 900 / 0.5

⇒ s = 1800 m

Therefore the distance travelled by the train before stopping is 1800 m (or) 1.8 km.