A motor bike running at 90, km h ^{-1} is slowed down to 18 ,km h^{-1} in 2.5, s . Calculate (i) acc

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7. MOTION

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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

Option A

Option B

Option C

Option D

Solution

$u=90 km h ^{-1}=25 m s ^{-1} ; v=18 km h ^{-1}=5 m s ^{-1}$

$t=2.5 s ; a=? ; S =?$

$(i)$ 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 $S=\frac{600}{16}=37.5 m$

Standard 9

Science

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