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

Question

A car moving with speed v on a straight road can be stopped with in distance d on applying brakes. If same car is moving with speed 3v and brakes provide half retardation, then car will stop after travelling distance (A) 6 d (B) 3 d (C) 9 d (D) 18 d

Tardigrade Admin · Accepted Answer

ds=(u2/2 a) ds propto u2 ⇒ (ds/ds')-(u2/u'2) u'=3 v u=v So, (ds/ds')=(v2/9 v2) ⇒ ds'=9 ds ⇒ ds'=9 d

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

6 d

3 d

9 d

18 d

$d_{s} = \frac{u ^{2}}{2 a}$
$d_{s} \propto u^{2} \Rightarrow \frac{d _{s}}{d _{s}^{'}} - \frac{u ^{2}}{u ^{^{'} 2}}$
$u^{'} = 3 v$
$u = v$
So, $\frac{d _{s}}{d _{s}^{'}} = \frac{v ^{2}}{9 v ^{2}}$
$\Rightarrow d_{s}^{'} = 9 d_{s}$
$\Rightarrow d_{s}^{'} = 9 d$

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

6 d

3 d

9 d

18 d

ds​=2au2​ ds​∝u2⇒ds′​ds​​−u′2u2​ u′=3v u=v So, ds′​ds​​=9v2v2​ ⇒ds′​=9ds​ ⇒ds′​=9d

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

$d_{s} = \frac{u ^{2}}{2 a}$
$d_{s} \propto u^{2} \Rightarrow \frac{d _{s}}{d _{s}^{'}} - \frac{u ^{2}}{u ^{^{'} 2}}$
$u^{'} = 3 v$
$u = v$
So, $\frac{d _{s}}{d _{s}^{'}} = \frac{v ^{2}}{9 v ^{2}}$
$\Rightarrow d_{s}^{'} = 9 d_{s}$
$\Rightarrow d_{s}^{'} = 9 d$