Can deceleration of car exceed ' $g$ '?

yes

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depends upon friction coefficient

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none

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Solution

The correct option is B no
Let $a$ be the deceleration of car.
Here, $v = 0, u = 60 k m / h r = \frac{60 \times 1000}{3600} = 16.67 m / s, S = 15 m$
by using, $v^{2} = u^{2} - 2 a S$ we will get,
$0 = (16.67)^{2} - 2 a (15)$
or $a = \frac{277.8}{30} = 9.3 m / s^{2}$
Thus $a$ can not exceed $g$ .

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Can deceleration of car exceed 'g'?

The correct option is B noLet a be the deceleration of car.Here, v=0,u=60km/hr=60×10003600=16.67m/s,S=15mby using, v2=u2−2aS we will get,0=(16.67)2−2a(15)or a=277.830=9.3m/s2 Thus a can not exceed g.

Can deceleration of car exceed ' $g$ '?

The correct option is B no
Let $a$ be the deceleration of car.
Here, $v = 0, u = 60 k m / h r = \frac{60 \times 1000}{3600} = 16.67 m / s, S = 15 m$
by using, $v^{2} = u^{2} - 2 a S$ we will get,
$0 = (16.67)^{2} - 2 a (15)$
or $a = \frac{277.8}{30} = 9.3 m / s^{2}$
Thus $a$ can not exceed $g$ .