Question

Consider a car moving on a straight roads with a speed of $100m/s$. The distance at which car can be stopped is $[{\mu }_x=0.5]$

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

Answer 1

The correct option is D $1000m$

When a car stopped by friction its retarding force = $\mu mg$

so $\mu mg=ma$

$a=\mu g$

So, equation $v^2=u^2-2as$

(a is negative because its retarding)

Final velocity v = 0

$u^2=2as$, where u = initial velocity

$S=\frac{4^2}{2a}$

$S=\frac{4^2}{2\mu g}$

$8=\frac{100\times 100\times 2}{20\times 1}=1000m$

Hence option (D) is correct