A jet lands on a runway at 110 m/s [forward]. When stopping, the jet can accelerate at 6.2 m/s2 [backward]. K/U T/I A (a) Calculate the minimum time for the jet to stop. [ans: 18 s] (b) What is the minimum safe length for this runway? [ans: 9.8 3 102 m] (c) Explain why the runway should be much longer than the minimum safe length.

a) Minimum time for the jet to stop = 18 secs approximately

b)  Minimum safe length for the runway = 975.6 m

c) The actual stopping distance would vary based on multiple factors like speed of the aircraft at the time of landing, weather conditions, change in friction due to wetness of the surface etc.. Hence the runway should be much longer than the minimum safe distance

Step-by-step explanation:

Step 1 :

Given,

Initial velocity of the jet = 110 m/s

Acceleration of the jet = - 6.2 m/s² [negative because its backward acceleration]

Final velocity = 0 [because the jet has to stop]

Step 2 :

a)

The minimum time for the jet to stop can be computed using the formula

v = u + at

where v represents the final velocity = 0

u = initial velocity = 110

a  = -6.2

Substituting,

110 - 6.2t = 0

t = -110 ÷ -6.2 =  17.74 secs which is approximately 18 secs

Step 3 :

b)

The minimum safe length for this runway can be computed using the formula

s = ut + at²

where s is the distance.

Substituting ,

s= 110× 18 +   (-6.2) ×18× 18

= 1980 - 1004.4 = 975.6 m

Step 4:

Answer :

a) Minimum time for the jet to stop = 18 secs approximately

b)  Minimum safe length for the runway = 975.6 m

c) The actual stopping distance would vary based on multiple factors like speed of the aircraft at the time of landing, weather conditions, change in friction due to wetness of the surface etc.. Hence the runway should be much longer than the minimum safe distance

1e+27

step-by-step explanation:

ab will be equal to