Answer:
Final
velocity, v = 0,
(\[\because
\]Car stops after applying the brakes)
\[a=-6\,m/{{s}^{2}},t=2s\]
From
first equation of y = u + at
or\[u=\upsilon
-at=0-(-6)(2)=12\,\text{m/s}\]
\[s=ut+\frac{1}{2}a{{t}^{2}}=(12)(2)+\frac{1}{2}(-6){{(2)}^{2}}\]
= 12
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