Solution) (1) v = 75 m/s
a = 1.3 m/s^2
L = 2.5 km
v^2 = u^2 + 2aS
(75)^2 = 0 + 2(1.3)(S)
S = (75^2)/(2.6)
S = 2,163.5 m
S = 2.163 Km
Since S < L , so the jet can safely use this runway
* kindly post remaining questions in next post
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