Question

if the velocity of a car is decreased by 50% then how much % decreased in stopping distance?​

Answer 1

let the initial velocity be'v' and stopping distance be 's' and retardation of brake be 'a'

• according to question
• 2as = v^2 - u^2
• 2as = 0 - v^2 _ equation 1

now if velocity is 1/2

so u =v/2

now distance

2(a)(s) = v^2 - u^2

2as = (v/2)^2

2as = (v^2)/4

subtitute value of eq 1

=1/4

so distance will become 1/4 times