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(N/A) Case $1$: Initial velocity $u = 90 \, km \, h^{-1} = 90 	imes \frac{5}{18} = 25 \, m \, s^{-1}$.Final velocity $v = 54 \, km \, h^{-1} = 54 	i

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(N/A) Case $1$: Initial velocity $u = 90 \, km \, h^{-1} = 90 	imes \frac{5}{18} = 25 \, m \, s^{-1}$.Final velocity $v = 54 \, km \, h^{-1} = 54 	imes \frac{5}{18} = 15 \, m \, s^{-1}$.Distance $S = 40 \, m$.Using the equation $v^2 - u^2 = 2aS$:$(15)^2 - (25)^2 = 2 	imes a 	imes 40$$225 - 625 = 80a$$-400 = 80a \implies a = -5 \, m \, s^{-2}$.Case $2$: The bike comes to rest from $90 \, km \, h^{-1}$ $(u = 25 \, m \, s^{-1})$,so $v = 0$.$(i)$ To find total time $t$,use $v = u + at$:$0 = 25 + (-5)t$$5t = 25 \implies t = 5 \, s$.$(ii)$ To find total distance $S$,use $v^2 - u^2 = 2aS$:$0^2 - (25)^2 = 2 	imes (-5) 	imes S$$-625 = -10S \implies S = 62.5 \, m$.

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