Solve the following :

Speed of car $=54 \times 18=15 \mathrm{~m} / \mathrm{s}$

Distance travelled during reaction time

$\mathrm{S}_{1}=\mathrm{v}^{\mathrm{X}} \mathrm{t}$

$=15 \times 0.2=3 \mathrm{~m}$

When brakes are applied

$\mathrm{u}=15 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-6 \mathrm{~m} / \mathrm{s}^{2} ; \mathrm{v}=0 \mathrm{~m} / \mathrm{s}$

$v^{2}=u^{2}+2 a s$

$0^{2}=(15)^{2}+2(-6) S_{2}$

$\mathrm{S}_{2}=18.75 \mathrm{~m}$

Total distance $=\mathrm{S}_{1}+\mathrm{S}_{2}$

$=3+18.75$

$=21.75 \mathrm{~m}$

$\approx 22 \mathrm{~m}$