If a car moves with velocity 'u' and its maximum retardation is 'a', its minimum stopping distance is proportional to

a)u

b)u²

c)u³

d)u^1/2

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Solution

The correct answer is option B

As we know, by using 3rd equation of motion we can say,
v²=u²+2as
As the body retards, resulting the body to stop final once its energy drains out. v=0
$0 = u^{2} + 2 (- a) (s)$
$s = \frac{u^{2}}{2 a}$
$s α u^{2}$

$∴\therefore$ s is proportional to u²

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If a car moves with velocity 'u' and its maximum retardation is 'a', its minimum stopping distance is proportional to a)u b)u2 c)u3 d)u1/2

The correct answer is option BAs we know, by using 3rd equation of motion we can say, v²=u²+2asAs the body retards, resulting the body to stop final once its energy drains out. v=00=u2+2(−a)(s) s=u22asαu2 ∴\therefore ∴ s is proportional to u²

If a car moves with velocity 'u' and its maximum retardation is 'a', its minimum stopping distance is proportional to

a)u

b)u²

c)u³

d)u^1/2

The correct answer is option B

As we know, by using 3rd equation of motion we can say,
v²=u²+2as
As the body retards, resulting the body to stop final once its energy drains out. v=0
$0 = u^{2} + 2 (- a) (s)$
$s = \frac{u^{2}}{2 a}$
$s α u^{2}$

$∴\therefore$ s is proportional to u²