The brakes of a car moving at $20 m / s$ 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.

2.22 m

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11.35 m

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22.2 m

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4.54 m

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Solution

The correct option is C $22.2 m$
Acceleration due to friction (retardation) (a) = $μ g = - 0.9 \times 10 = - 9 m / s^{2}$
Final velocity, $v = 0 m / s$
Initial velocity, $u = 20 m / s$
Applying formula; $v^{2} = u^{2} + 2 a s$
Where $s$ is the distance travelled by car before it comes to rest.
$\Rightarrow 0^{2} = 20^{2} + 2 (- 9) s$
$\Rightarrow 18 s = 400$
$\Rightarrow s = 22.2 m$

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The correct option is C 22.2 mAcceleration due to friction (retardation) (a) = μg=−0.9×10=−9 m/s2 Final velocity, v=0 m/s Initial velocity, u=20 m/s Applying formula; v2=u2+2as Where s is the distance travelled by car before it comes to rest. ⇒02=202+2(−9)s ⇒18s=400 ⇒s=22.2 m