Question

When the speed of a car is v, the minimum

distance over which it can be stopped is s. If the

speed becomes nv, what will be the minimum

distance over which it can be stopped during same

time?

(a) s/n (b) ns (c)

2

s/n

(d)

2

ns​

Answer 1

Given:

Let the speed of car be v

Minimum distance cover=s

final speed =0

Acceleration=?

Solution

From third equation of motion:

v^2-u^2=2as

0^2 -v^2=2as

a= - v^2 /2a

If speed becomes nv then minimum distance cover which it can be stopped during same retardation ( -v^2/2s)

From third equation of motion:

s= -v^2/2a

where v= nv

a= -v^2/2a

s= - (nv)^2/ 2x(-v^2)/2s

s=n^2 s

So minimum distance with same retardation is n square s .