Question

10. A car is moving at a certain speed. The minimum
distance over which it can be stopped is x. If the
speed of the car is doubled, what will be the minimum
distance over which the car can be stopped during the
same time?
(1) 4x
(2) 2x
(3) x/2
(4) x/4​

Answer 1

Answer:

Given:

Let the speed of car be v

Minimum distance cover=x

final speed =0

Acceleration=?

From third equation of motion:

v^2-u^2=2ax

0^2 -v^2=2ax

a= - v^2 /2a

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

From third equation of motion:

x'= -v^2/2a

where v= 2v

a= -v^2/2a

x'= - (2v)^2/ 2x(-v^2)/2x

x'=4 x

So minimum distance with same retardation is n square s .