Question

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

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

Answer 1

Answer: (D)

$u=42 \times \frac{5}{18}=11.66\, m / s$
and $v=0$ So, $s_{1}=6 m$
when, $u=90 \times \frac{5}{18}=25\, m / s$
and $v=0 .$
Then, $s_{2}=?$
For first case $\Rightarrow v^{2}=u^{2}+2 a s_{1}$
$0=(11.66)^{2}+2 a \times 6$
$\left(s_{1}=6 m \right)$
$a=\frac{-11.66 \times 11.66}{12}$
$a=-11.33\, m / s ^{2}$
For second case $v^{2}=u^{2}+2 a s_{2}$
$0=(25)^{2}+2(-11.33) \times s_{2}$
$s_{2}=\frac{625}{2 \times 11.33}=27.5\, m$