Question

Consider a car moving on a straight road with a speed of $100m{s}^{-1}$. The distance at which car can be stopped is $[{\mu }_k=0.5]$

• A. $800m$
• B. $1000m$
• C. $100m$
• D. $400m$

Answer 1

The correct option is B $1000m$

From third equation of Motion, we have
$v^2=u^2+2as$
Given, $v=0$ [car is stopped]
As friction provide the retardation
$a=-\mu g,u=100m{s}^{-1}$
$\therefore (100{)}^2=2\mu gs$
$\Rightarrow s=\frac{100\times 100}{2\times 0.5\times 10}$
$=\frac{100\times 100}{5\times 2}=1000m$