Question

when a speed of a car is v,the minimum distance over which it it can stopped is s.If the speed becomes nv,what will be the minimum distance over which it can be stopped during same retardation?

Answer 1

Given:

Let the speed of car be v
Minimum distance cover=s
final speed =0
Acceleration=?

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 .

Answer 2

Answer:

Pls refer to the solution in the image attached

Explanation:

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