UPSEE Solved Model Paper 6

Here, we have to find out how the minimum distance required to stop a car changes, when the speed of the car is reduced to 2/5th.
We know that,
According to work - energy principle,
Work done in stopping = KE
ie, F.S = ½ mv2
Where,
F - Force
S - displacement
m - mass
v - velocity
If F & ½ m are constants,
S α v2
When the velocity is reduced (2/5)th of the initial one,
S' α (

2

5

v)2
i.e.,

S′

S

=

4

25

S' =

4

25

S
That is, the minimum distance required to stop the car becomes (4/25)th times of the initial one.