$A$ motorbike running at a speed of $90\, km h^{-1}$ slows down to $18\, km h^{-1}$ in $2.5\, s$. Calculate:
$(i)$ Acceleration
$(ii)$ Distance covered during the deceleration

(N/A) Initial velocity $u = 90\, km h^{-1} = 90 \times \frac{5}{18} = 25\, m s^{-1}$.
Final velocity $v = 18\, km h^{-1} = 18 \times \frac{5}{18} = 5\, m s^{-1}$.
Time $t = 2.5\, s$.
$(i)$ Using the first equation of motion $v = u + at$:
$5 = 25 + a \times 2.5$
$5 - 25 = 2.5a$
$-20 = 2.5a$
$a = \frac{-20}{2.5} = -8\, m s^{-2}$.
Thus,the acceleration is $-8\, m s^{-2}$ (which indicates retardation).
$(ii)$ Using the third equation of motion $v^2 - u^2 = 2aS$:
$(5)^2 - (25)^2 = 2 \times (-8) \times S$
$25 - 625 = -16S$
$-600 = -16S$
$S = \frac{600}{16} = 37.5\, m$.
Thus,the distance covered is $37.5\, m$.