Question

A train is travelling at a speed of 90kmPh. Brakes are applied so as
to produce a uniform acceleration of -0.5ms-2. Find the distance travelled
by the train before it comes to rest.

Answer 1

Answer :-

Given:-

Initial velocity (u) = 90 Km/h

Final velocity (v) = 0 m/s

Acceleration (a) = -0.5 m/s²

Converting Km/h to m/s:-

⟹ 90 × 5/18 m/s

⟹ 25 m/s

Using third equation of motion:-

⟹ v² = u² + 2as

Hence:-

⟹ 0² = 25² + 2(-0.5)s

⟹ 0 = 625 + (-1)s

⟹ 0 = 625 - s

⟹ s = 625 m

∴ The distance travelled by the train before it comes to rest is 625 m.

Answer 2

QUESTION:-

A train is travelling at a speed of 90kmPh. Brakes are applied so as

to produce a uniform acceleration of -0.5ms². Find the distance travelled

by the train before it comes to rest.

EXPLANATION:-

• Initial velocity(u)=0 m/s
• Final velocity(v)=90 km/hr=90×5/18=25 m/s
• Acceleration(a)=-0.5 m/s²

Now  let's use 2nd equation of motion

v² = u² + 2as

 0² = 25² + 2(-0.5)s

0 = 625 + (-1)s

 0 = 625 - s

s = 625 m

Thus,

distance covered is 625 metres

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