Question

The brakes of a car moving at $20m/s$ along a horizontal road are suddenly applied and it comes to rest after travelling some distance. If the coefficient of friction between the tyres and the road is $0.90$ and it is assumed that all four tyres behave identically, find the shortest distance the car would travel before coming to a stop.

• A. $2.22m$
• B. $11.35m$
• C. $22.2m$
• D. $4.54m$

Answer 1

The correct option is C $22.2m$

Acceleration due to friction (retardation) (a) $=$ - $\mu g=-0.9\times 10$=$-9m/s^2$
Final velocity, $v=0m/s$
Initial velocity, $u=20m/s$
Applying formula: $v^2=u^2+2as$
Where $s$ is the distance traveled by car before it come to rest.
$\Rightarrow 0^2={20}^2+2(-9)s$

$\Rightarrow 18s=400$
$\Rightarrow s=22.2m$