Question

A train was moving at the rate of 54 kmh-1 when brakes were applied. It came to rest

within a distance of 225m. Calculate the retardation produced in the train.​

Answer 1

Given:-

• Final velocity of train = 0m/s

• Initial velocity of train = 54km/h → 15m/s

• Distance travelled = 225m

To Find:-

• The Retardation of train

Formulae used:-

• v² - u² = 2as

Where,

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

Now,

→ v² - u² = 2as

→ (0)² - (15)² = 2 × a × 225

→ -225 = 450a

→ a = -225/450

→ a = -0.5m/s²

Hence, The Retardation Produced in train is -0.5m/s².

Answer 2

Explanation:

Given:-

final velocity of train=v=0m/s

initial velocity of train=u=54km/h

${:}\longrightarrow$15m/s

Distance travelled =s=225m

To find:-

The retardation of train

Solution:-

Formula used:-

${:}\longrightarrow$${\large{\boxed{\sf{{v}^{2}-{u}^{2}=2as}}}}$

Where,

• s=distance
• u=initial velocity
• v=final velocity
• a=acceleration

Now,

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

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

${:}\longrightarrow$$-225=450a$

${:}\longrightarrow$$a={\frac{-225}{450}}$

${:}\longrightarrow$$a={\frac {-1}{2}}$

${:}\longrightarrow$${\underline{\boxed{\bf{a=-0.5m/s {}^{2}}}}}$

Hence,

The retardation produced in the train is $-0.5m/{s}^{2}$