Question

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A train is travelling with a speed of 72 Kmph. The driver applies brakes so that a uniform acceleration
of 0.2 ms is produced. The distance travelled by the train before it comes to rest is
a) 1 km
b) 10 km
C) 0.1 km​

Answer 1

Answer:

Apply formula v^2 = u^2 - 2as....(1)

Here v = 0, a = 0.2 m/s^2

u = 72km/hr = 72×1000m/3600s = 20m/s

Put the values of u, v, a, in (1) we get

20×20 = 2 × 0.2 ×s

s = 20×20/(2×0.2) = 1000m 1km Ans

Answer 2

Answer

• Distance travelled by the train will be 1 km [Option A]

Explanation

Given

• Speed = 72 km/h
• Acceleration = 0.2 m/s²

To Find

• Distance travelled by the train before it stops

Solition

• So here w shall first convet the speed from km/h to m/s

→ 72 × 5/18

→ 4×5

→ Initial Velocity = 20 m/s

✭ Distance travelled by the train

→ v²-u² = 2as

• v = Final Velocity = 0 m/s
• u = Initial Velocity = 20 m/s
• a = Acceleration = 0.2 m/s²
• s = Distance = ?

→ 0²-20² = 2 × 0.2 × s

→ 0-400 = 0.4s

→ -400/0.4 = s

→ s = -1000 m

[Ignoring negative sign]

→ s = 1000 m

Now 1000 m will be 1 km

→ Distance = 1 km