A motor car slows down from $72\, km h ^{-1}$ to $36\, km h^{-1}$ over at distance of $25\, m$. If the brakes are applied with the same force calculate $(i)$ total time in which car comes to rest $(ii)$ distance travelled by it.
A motor car slows down from $72\, km h ^{-1}$ to $36\, km h^{-1}$ over at distance of $25\, m$. If the brakes are applied with the same force calculate $(i)$ total time in which car comes to rest $(ii)$ distance travelled by it.
Case 1: $u=72 km h ^{-1}=20 m s ^{-1} ; v=36 km h ^{-1}$
$=10 m s ^{-1} ; S =25 m ; a=?$
Applying $\quad v^{2}-u^{2}=2 a S$
$(10)^{2}-(20)^{2}=2 \times a \times 25$
$-300=50 a$
or $a=-6 m s ^{-2}$
Case $2$ : $u=72 km h ^{-1}=20 m s ^{-1} ; v=0 ; S =?$
$t=? ; a=-6 m s ^{-2}$
Applying $v^{2}-u^{2}=2 a S$
$(0)^{2}-(20)^{2}=2 \times(-6) \times S$
$S=\frac{-400}{-12}=33.33 m$
Applying $\quad=u+a t$
$0=20-6 \times t$
$t=\frac{20}{6}=3.33 s$
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