A motor bike running at $90 \,km h ^{-1}$, is slowed down to $54 \,km h^{-1}$ by the application of brakes, over a distance of $40\, m$. If the brakes are applied with the same force, calculate $(i)$ total time in which bike comes to rest $(ii)$ total distance travelled by bike.

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Case $1$ : $u=90 km h ^{-1}=(-) 25 ms ^{-1} ; v=54 km$

$h^{-1}=15 m \cdot s ^{-1} ; S =40 m ; a=?$

Applying $v^{2}-u^{2}=2 a S$

$(15)^{2}-(25)^{2}=2 \times a \times 40$

$80 a=-400$

$a=-5 m s ^{-2}$

Case $2$ : $u=90 km h ^{-1}=25 m s ^{-1} ; v=0 ; S =?$

$a=-5 m s ^{-2} ; t=?$

Applying $v^{2}-u^{2}=2 a S$

$(0)^{2}-(25)^{2}=2 \times(-5) \times S$

$S=\frac{-625}{-10}=62.5 m$

Applying $\quad v=u+a t$

$0=25-5 \times t$

or $5 t=25$ or $t=5 s$

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