Question

When a train is stopped by applying break it stops after travelling a distance of 50 m. If speed of train is doubled and same retarding force is applied then it stops after travelling a distance of:

• A. $50m$
• B. $100m$
• C. $200m$
• D. $400m$

Answer 1

The correct option is C 200m

Let retardation is ‘a’.

Here train stops therefore final velocity , v=0

Using third equation of motion:
v²=u²+2as1

⇒ 0=u²−2as1
⇒ s1=u²/2a.....(i)
When speed of train is doubled, let train travel s2 distance
so s2=(2u)²/2a...(ii)

{given:s1=50m}
s2=4(u²/2a)

Using equation (i) we get

⇒s2=4*s1=4×50=200m