A motor car slows down from 72, km h ^{-1} to 36, km h^{-1} over at distance of 25, m . If brakes ar

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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.

Option A

Option B

Option C

Option D

Solution

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 $\quad=u+a t$

$0=20-6 \times t$

$t=\frac{20}{6}=3.33 s$

Standard 9

Science

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