An aeroplane lands at $216 \ km \ h^{-1}$ and stops after covering a runway of $2 \ km$. Calculate the acceleration and the time in which it comes to rest.

(N/A) Given:
Initial velocity $u = 216 \ km \ h^{-1} = 216 \times \frac{5}{18} \ m \ s^{-1} = 60 \ m \ s^{-1}$.
Final velocity $v = 0 \ m \ s^{-1}$ (as it comes to rest).
Distance covered $S = 2 \ km = 2000 \ m$.
$(i)$ To find acceleration $(a)$,we use the equation $v^{2} - u^{2} = 2aS$:
$(0)^{2} - (60)^{2} = 2 \times a \times 2000$
$-3600 = 4000 \times a$
$a = -\frac{3600}{4000} = -0.9 \ m \ s^{-2}$.
$(ii)$ To find the time $(t)$,we use the equation $v = u + at$:
$0 = 60 + (-0.9) \times t$
$0.9t = 60$
$t = \frac{60}{0.9} \approx 66.67 \ s$.