Question

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Answer 1

Given :

• A train is moving at 108 km / hr .after applying brake the speed of train reduce to 72 km/hr within 20 second.

To Find :

• Acceleration (a) = ?
• Distance (s) = ?

Solution :

• ⌬ Final Velocity (v) = 72 km/hr
• ⌬ Initial Velocity (u) = 108 km/hr
• ⌬ Time taken (t) = 20 seconds

★ First of all convert initial velocity from km/hr to m/s :-

• ⇒Initial Velocity (u) = 108 km/hr
• ⇒Initial Velocity (u) = 108 × 5/18
• ⇒Initial Velocity (u) = 6 × 5
• ⇒Initial Velocity (u) = 30 m/s

Hence,the initial velocity of the train is 30 m/s.

★ Now, convert the final velocity from km/hr to m/s :-

• ⇒Final Velocity (v) = 72 km/hr
• ⇒Final Velocity (v) = 72 × 5/18

1. ⇒Final Velocity (v) = 4 × 5

• ⇒Final Velocity (v) = 20 m/s

Hence,the final velocity of the train is 20 m/s.

★ Finding the acceleration of the train by using first equation of motion :

• → v = u + at
• → 20 = 30 + a × 20
• → 20 - 30 = 20a
• → -10 = 20a
• → a = -10 ÷ 20
• → a = -1 ÷ 2
• → a = -0.5 m/s²

• Hence,the acceleration of the train is -0.5 m/s².

★ Now, let's calculate the distance travelled by the train by using third equation of motion :

• → v² - u² = 2as
• → (20)² - (30)² = 2 × -0.5 × s
• → 400 - 900 = -1s
• → -500 = -1s
• → s = -500 ÷ -1
• → s = 500 ÷ 1

→ s = 500 m

Hence,the distance travelled by the train is 500 m.