Question

train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of –0.5 m s-2. Find how far the train will go before it is brought to rest.​

Answer 1

Answer:

find velocity.

velocity=displacement\time

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Answer 2

Given :-

Initial velocity = 90 km/h

Terminal velocity = 0 m/s²

Acceleration = -0.5 m/s⁻²

To Find :-

How far the train will go before it is brought to rest.​

Solution:-

We know that,

• u = Initial velocity
• t = Time
• v = Final velocity
• a = Acceleration

Using the third equation of motion,

$\underline{\boxed{\sf Third \ equation= \dfrac{v^2-u^2}{2} }}$

Given that,

Initial velocity (u) = 90 km/h

Final velocity (v) = 0 m/s²

Acceleration (a) = -0.5 m/s⁻²

Substituting their values,

s = (0²-25²)/2(-0.5)

s = 625 m

Therefore, the train must travel 625 meters before it is brought to rest.​