Question

A train is travelling at a speed of 90km/ h breaks are applied to produce a uniform acceleration of -0.5m/s find how far the train will go before the it is brought to rest

Answer 1

Given:-

• Initial velocity,u = 90 km/h

• Acceleration,a = -0.5 m/s²

• Final velocity,v = 0 m/s

To be calculated:-

Find how far the train will go before the it is brought to rest.

Formula used:-

v² = u² + 2as

Solution:-

We need to convert 90 km/h into m/s

1 km = 1000 m

90 km = 90000 m

And,

1 hour = 60 × 60

= 3600 seconds

So,

Initial velocity = 90 km/h

= 90000/3600

= 25 m/s

Now,

From the third equation of motion:

v² = u² + 2as

★ Putting the values in the above formula,we get;

( 0 )² = ( 25 )² + 2 × - 0.5 × s

⇒ 0 = 625 + 2 × - 5/10 × s

⇒ 0 = 625 + ( - 1 ) × s

⇒ 1 × s = 625

⇒ s = 625/1

⇒ s = 625 m

Hence,the distance covered by train is 625 m.

Answer 2

Answer :

• 625 meters

Given :

• A train is travelling at a speed of 90 km/ h breaks are applied to produce a uniform acceleration of -0.5 m/s
• Initial velocity , u = 90 km/h = 90*(5/18) = 25 m/s
• Final velocity , v = 0 m/s [ Since brought to rest ]
• acceleration , a = - 0.5 m/s²

To Find :

• How far the train will go before brought into rest
• Distance , s = ?

Solution :

Use 3 rd equation of motion .

⇒ v² - u² = 2as

⇒ 0² - 25² = 2(-0.5)s

⇒ -625=-s

⇒ s = 625 m

So the train will go 625 m before brought into rest

More Info :

• - ve sign of acceleration shows retardation or deceleration .
• Retardation is nothing but acceleration but opposite direction .