A motor bike running at $90\, km h ^{-1}$ is slowed down to $18 \,km h^{-1}$ in $2.5\, s$. Calculate
$(i)$ acceleration
$(ii)$ distance covered in the time it slows down
A motor bike running at $90\, km h ^{-1}$ is slowed down to $18 \,km h^{-1}$ in $2.5\, s$. Calculate
$(i)$ acceleration
$(ii)$ distance covered in the time it slows down
$u=90 km h ^{-1}=25 m s ^{-1} ; v=18 km h ^{-1}=5 m s ^{-1}$
$t=2.5 s ; a=? ; S =?$
$(i)$ Applying $v=u+a t$
$5=25+a \times 2.5$
$-20=2.5 \times a$
Or $a=\frac{-20}{2.5}=-8 m s ^{-2}$
$(ii)$ Applying $v^{2}-u^{2}=2 a S$
$(5)^{2}-(25)^{2}=2 \times(-8) \times S -600=-16 S$
or $S=\frac{600}{16}=37.5 m$
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