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) 2s/n
(d) 2ns​

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 .

thank you...