Question

A car moving with a speed of $40\text{km/h}$ can be stopped by applying brakes after at least $2\text{m}$. If the same car is moving with a speed of $80\text{km/h}$, what is the minimum stopping distance

• A. $8\text{m}$
• B. $2\text{m}$
• C. $4\text{m}$
• D. $6\text{m}$

Answer 1

The correct option is A $8\text{m}$

For $1^{st}$ case:
$u=40\text{km/h}=(40\times \frac{5}{18})\text{m/s}$
From third equation of motion we have,
$v^2-u^2=2as$
$v^2=u^2+2as$
$0={(40\times \frac{5}{18})}^2+2a\left(2\right)$
$a=-\frac{1}{4}{\left(\frac{100}{9}\right)}^2......\left(1\right)$

For $2^{nd}$ case:
$u=80\text{km/h}=(80\times \frac{5}{18})\text{m/s}$
From third equation of motion we have,
$v^2=u^2+2as$
$0={(80\times \frac{5}{18})}^2+2\times a\times s$
$2\times a\times s=-{\left(\frac{200}{9}\right)}^2$
$-2\times \frac{1}{4}\times {\left(\frac{100}{9}\right)}^2\times s=-{\left(\frac{200}{9}\right)}^2$
on solving we get
$s=8\text{m}$