Question

A trai is travelling at a speed of 90km/hr.Brakes are applied so as to produce a uniform acceleration of -0.5 m/s2.Find how far the train will go before it is bought to rest..give ans in easy step .............................

Answer 1

Given :-

• Initial speed (u) = 90km/h
• Final velocity (v) = 0 (brakes applied)
• Acceleration (a) = - 0.5m/s²

To find :-

• Distance travelled (s)

Solution :-

• Initial velocity = 90km/h

→ u = 90 × 5/18

→ u = 25m/s

According to the third equation of motion

→ v² = u² + 2as

→ (0)² = (25)² + 2 × (-0.5) × s

→ 0 = 625 - s

→ s = 625 m

Hence,

• Distance travelled by train is 625m

Extra Information :-

• v = u + at (First equation of motion)
• s = ut + ½ at² (second equation of motion)
• S.I unit of distance is meter

Answer 2

$\star\:\:\:\bf\large\underline\blue{Given:—}$

☛︎ Acceleration of the train, a = –0.5 m/s²

☛︎ Speed, v = 90 km/h = 25 m/s

$\star\:\:\:\bf\large\underline\blue{To\:find:—}$

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

$\star\:\:\:\bf\large\underline\blue{Question:—}$

Using the equation of motion ,

$\boxed{\bf{\red{v=u+at}}}$

Put the value into the equation

↬0 = 25- 0.5 × t

↬25 = 0.5 × t

↬t = 25/0.5

↬t = 50 seconds

Again using the equation of motion,

$\boxed{\bf{\red{s=ut+\frac{1}{2}at²}}}$

Put the value into the equation,

Where, s is distance travelled before stop

➝ S = 25 × 50 - 1/2 × 0.5 × (50)² m

➝ S = 625 m

Hence, the train will go before it is brought to rest is 625 m

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