Question

The bus moving with a speed of $42km/h$ is brought to a stop by brakes after $6m$. If the same bus is moving at a speed of $90km/h$, then the minimum stopping distance is

• A. $15.48m$
• B. $18.64m$
• C. $22.13m$
• D. $27.55m$

Answer 1

Answer: (D)

$u=42\times \frac{5}{18}=11.66m/s$
and $v=0$ So, $s_1=6m$
when, $u=90\times \frac{5}{18}=25m/s$
and $v=0.$
Then, $s_2=?$
For first case $\Rightarrow v^2=u^2+2a{s}_1$
$0=(11.66{)}^2+2a\times 6$
$\left(s_1=6m\right)$
$a=\frac{-11.66\times 11.66}{12}$
$a=-11.33m/s^2$
For second case $v^2=u^2+2a{s}_2$
$0=(25{)}^2+2(-11.33)\times s_2$
$s_2=\frac{625}{2\times 11.33}=27.5m$