Question

A car moving with speed v on a straight road can be stopped with in distance d on applying brakes. id same car is moving with speed $3$v and brakes provide half retardation, then car will stop after travelling distance.

Answer 1

assuming retardation is a

initial velocity = v

Final velocity = 0

Use, formula , V² = U² + 2aS

Here , V is final velocity , U is intial velocity , a is acceleration and S is the distance.

now, 0 = v² + 2(-a)d [ because a is retardation so, we took -a ]

v² = 2ad

a = v²/2d

So, retardation in 2nd case , a' = a/2 = v²/4d

Now, again use, S = u²/2A

Here, u = 3v and A = retardation = v²/4d

Now, S = (3v)²/2(v²/4d)

S = 9 × 2 d = 18d

Hence, answer is 18d