Question

when speed of a car v the minimum distance over which it over which it can stopped is x . if the speed become n×v , what will be the minimum distance over which it can stoped?

a) x/n. b) n×x c) x/n^2. d). n^2/x.

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Answer 1

v²=u²⁺2as
v²=2ax
(nv)²=2as
s=n²x

Answer 2

Using the third equation of motion:
${v}^{2} = {u}^{2} + 2as$
We know the car eventually stops, so v = 0.

Case 1: When speed is v.
$0 = {v}^{2} - 2ax$
Acceleration is negative, that's why the car will stop.
$x = \frac{ {v}^{2} }{2a}$
Case 2: When speed is nv.
Let the distance over which it can stop be s.
$0 = {n}^{2} {v}^{2} - 2as$
s=n^2(v^2/2a)
Putting (v^2/2a=x)
s=n^2x