A car slows from 23.69 m/s to rest in 4.44 s. How far did it travel in this time?
Answer
Using
v=u+at
0=23.69−a×4.44a=5.35m/s20=23.69-a\times4.44\\a=5.35m/s^20=23.69−a×4.44a=5.35m/s2
So travelled distance
S=v2−u22a=0−23.692−2×5.35=52.45mS=\frac{v^2-u^2}{2a}\\=\frac{0-23.69^2}{-2\times5.35}\\=52.45mS=2av2−u2=−2×5.350−23.692=52.45m
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