if the velocity of a car is increased by 20%, then the minimum distance in which it can be stopped increases by

Question

Saksham Bansal · Answer

As if deacceleration in both cases is kept constant then as:-
$s= (v^2- u^2 )/ 2a$ and in both cases v=0 thus
s₁= -u²/2a and s₂= -(1.20u)²/2a thus change in stopping distance
(s₂-s₁)
thus increase in stopping distance= -1.44u²/2a – (-u²)/2a = -0.44u²/2a
thus distance increases by 0.44×100=44%

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if the velocity of a car is increased by 20%, then the minimum distance in which it can be stopped increases by

if the velocity of a car is increased by 20%, then the minimum distance in which it can be stopped increases by

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if the velocity of a car is increased by 20%, then the minimum distance in which it can be stopped increases by

if the velocity of a car is increased by 20%, then the minimum distance in which it can be stopped increases by

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